In a scientific experiment, whether in physics, chemistry, or another natural science discipline, it is very common to have to perform measurements and, with them, establish multiple operations that lead to various conclusions. In this way, it is important to recognize which factors produce or affect a phenomenon that we are studying.
What is a variable in science?
A variable is any aspect that can modify its value in an experiment.
Many phenomena studied during an experiment depend on various variables, which are necessary to obtain correct results in the laboratory. These are the important factors that we must take into account when experimenting and are defined as those data that can change in value during an experiment. It is necessary to pay attention to all possible variables that appear in an experiment.
Consequently, those variables that do not significantly affect the result or are of little importance should be eliminated (for example, we can cite the variable “age of the experimenter”: this variable is insignificant for studying the value of Earth’s gravity at a specific site, but it is significant for studying the incidence of an endemic disease in that site). When the time comes, each variable of the experiment should be modified to study the phenomenon.
In scientific research, variables are used to describe and measure the different factors that can influence the outcome of an experiment. There are three main types of variables: independent, dependent, and controlled. Each of these variables plays a unique role in the scientific process and helps researchers draw valid conclusions from their experiments. In this article, we will explore these three types of variables and provide examples of how they are used in scientific research.
Types of variables
There are three types of variables that we will detail below:
Independent Variables
Independent variables are the variables that the researcher manipulates or changes deliberately in an experiment. These variables are chosen by the researcher to test their hypothesis and see how they affect the dependent variable. For example, in an experiment testing the effects of caffeine on alertness, the independent variable is the amount of caffeine consumed by participants. The researcher can vary the amount of caffeine to see how it affects alertness, which is the dependent variable.
Another example of an independent variable is the age of participants in a study. Researchers may want to investigate how age affects cognitive function, so they deliberately select participants from different age groups and compare their performance on cognitive tasks. In this case, age is the independent variable.
Dependent Variables
Dependent variables are the variables that are measured or observed in an experiment. They are called “dependent” because their values depend on the independent variable. In our example of the effects of caffeine on alertness, alertness is the dependent variable. It is the variable that is affected by the amount of caffeine consumed by participants. The researcher measures alertness to determine the effect of caffeine on it.
Another example of a dependent variable is the amount of weight loss in a weight-loss program. Researchers may want to test the effectiveness of a particular diet or exercise program, so they measure the amount of weight lost by participants who follow the program. In this case, the amount of weight lost is the dependent variable.
Controlled Variables
Controlled variables are the variables that are kept constant in an experiment. They are used to ensure that any changes in the dependent variable are due to changes in the independent variable and not to other factors. For example, in an experiment testing the effects of caffeine on alertness, the temperature in the room could be a controlled variable. The researcher would keep the temperature constant throughout the experiment to ensure that it does not affect alertness.
Another example of a controlled variable is the type of measuring instrument used in an experiment. Researchers may want to investigate the effect of a particular treatment on blood pressure, so they use the same type of blood pressure cuff to measure blood pressure for all participants. This ensures that any changes in blood pressure are due to the treatment and not to differences in the measuring instrument.
Conclusion
In conclusion, independent, dependent, and controlled variables are important components of scientific research. Independent variables are the variables that are manipulated or changed deliberately in an experiment. Dependent variables are the variables that are measured or observed in an experiment and are affected by the independent variable. Controlled variables are the variables that are kept constant in an experiment to ensure that changes in the dependent variable are due to changes in the independent variable. Understanding these types of variables is essential for conducting valid and reliable scientific research.
In this Fluid Mechanics article, we will see drainage of a fluid and how to calculate the exit velocity of a fluid through a container. The interesting thing about this topic is that we can, applying the same logic and reasoning, handle cases where we have a liquid container with an opening at the bottom (such as a tank with a hole in its lower part, as shown in Figure 1) or a container that has a hole in its walls (as shown in Figure 2).
Figure 1: Drainage of a fluid.
Figure 2: Exit velocity in a pierced container.
Separemos nuestro estudio en ambos casos.
What is fluid mechanics?
Fluid mechanics is a branch of physics that deals with the study of the behavior of fluids (liquids and gases) at rest or in motion. It aims to describe how fluids move and interact with their surroundings.
Fluid mechanics is applied in a wide variety of fields, from civil and mechanical engineering to medicine and biology. Some of the topics studied in fluid mechanics include fluid dynamics (movement of fluids), fluid kinematics (velocity and acceleration of the fluid), fluid thermodynamics (heat transfer in fluids), and computational fluid mechanics (use of computational tools to simulate the behavior of fluids).
Exit velocity in a container with a hole (in side walls or bottom)
Figure 1 shows a container with a cross-sectional area A, filled with a fluid up to a height that we will call h. The space above the tank is at atmospheric pressure \( p_{1}\) , and the fluid exits through a tube with area \( A_{2}$\) .
Let’s try to obtain an expression for the exit velocity. We can locate point 1 on the surface of the liquid. There, the pressure will be called \( p_{1}\), and the fluid velocity will be \( v_{1}\) . Point 2 will be located at the tank outlet.
There, we will call \( v_{2}\) to the exit velocity, and its pressure will be denoted as \( p_{2}\) . We can set the level of point 2 as \( h=0\) , so the height of point 1 will be equal to h. Since area 1 is much larger than area 2, we can say that the fluid level in the tank will drop very slowly.
We could say that the velocity at point 1 is zero (practically, \( v_{1}=0\) ). Another important consideration is the following: The exit velocity will depend on several factors. But… which ones? To answer this, let’s apply the Bernoulli’s Principle.
We see that, of course, the velocity \( v_{2}\) will depend on the gravity and density of the fluid. What is not so evident and we must always keep in mind is that said velocity will depend on the pressure difference \( \Delta p \) between the upper and lower points, that is, it depends on \( \Delta p = p_{2}-p_{1}\), and also on the height h of the liquid.
At this point, we have a lot of information that will make it easy for us to work with the Bernoulli Equation. Let’s go step by step.
If we consider that both the top and bottom of the container are at atmospheric pressure (that is, \( p_{1}=p_{2}=p_{atm}\), so \( \Delta p =0\)), we can cancel out the pressures:
Rearranging (remembering that \( \frac{1}{\frac{1}{2}}=2\)) and eliminating \( \delta\) because it appears in both the numerator and denominator:
\(2\cdot g\cdot h_{1}=v_{2}^{2}\)
We obtain the square root to solve for the exit velocity:
\( v_{2}=\sqrt{2\cdot g\cdot h_{1}}\)
And that’s it! We have obtained the expression for the exit velocity of a fluid! Do you notice something similar? In free fall, the speed with which an object hits the ground presents the same expression, with h being the height from which said object is dropped. This result is known as Torricelli’s theorem and is valid not only for a tank that has an opening at the bottom but also for the sides of the container.
Flow rate of volume in a container with drain
Calculating the volume flow rate in a container that has a drain, whether it is at the bottom or on the side, is very simple, by applying Torricelli’s theorem and the Principle of Continuity.
We know that the flow rate must remain constant over time, meaning that the flow rate at the top point must be equal to the flow rate at the bottom point. Therefore:
\( C_{1}=C_{2}\)
As \( C=A\cdot v\), we can establish the relationship:
\( C= A_{2} \cdot v_{2}\)
Given that \( v_{2}=\sqrt{2\cdot g\cdot h_{1}}\) (as we deduced before), then:
\( C= A_{2} \cdot \sqrt{2\cdot g\cdot h_{1}}\)
Activities about drainage of a fluid
A tank has a capacity of 1000 liters and is filled with water. The tank drain has a diameter of 5 cm and a length of 50 cm. If the flow of water coming out of the drain is 2 liters per minute, how long will it take for the tank to empty completely?
A conical tank has a height of 5 meters and a base diameter of 4 meters. The drain of the tank is at the bottom and has a diameter of 5 cm. If the flow of water coming out of the drain is 1 liter per minute, what is the volume flow rate of water leaving the tank?
A tank has a capacity of 500 liters and is filled with a saline solution. The tank drain has a diameter of 2 cm and a length of 20 cm. If the concentration of salt in the solution is 0.1 g/ml and the flow of water coming out of the drain is 1 liter per minute, what is the flow rate of salt leaving the tank?
A cylindrical tank with a height of 2 meters and a diameter of 1 meter is filled with water up to a height of 1.5 meters. If there is a circular hole with a diameter of 3 centimeters at the bottom of the tank, at what rate will the water flow out of the hole? Assume that the velocity of the water at the hole is the same as that of the fluid at the surface of the tank.
A conical tank with a height of 5 meters and a base radius of 3 meters is filled with oil up to a height of 4 meters, and there is a circular hole with a diameter of 2 centimeters at the bottom of the tank. At what rate will the oil flow out of the hole? Assume that the velocity of the oil at the hole is the same as that of the fluid at the surface of the tank.
A rectangular tank has a height of 1 meter, a length of 2 meters, and a width of 0.5 meters. The tank is filled up to a height of 80 centimeters with water, and there is a rectangular hole with a width of 5 centimeters and a height of 10 centimeters in one of the side walls of the tank. If the velocity of the water at the hole is the same as that at the surface of the water in the tank, at what rate will the water flow out of the hole?
Answer key:
The volume flow rate of water exiting the drain is 2 liters per minute. If the tank has a capacity of 1000 liters, it will take 500 minutes (or 8 hours and 20 minutes) to completely empty. This can be calculated by dividing the tank capacity by the volume flow rate: 1000 L / 2 L/min = 500 min.
The volume of the conical tank can be calculated using the formula V = (1/3)πr²h, where r is the radius of the base and h is the height of the cone. In this case, the radius is 2 meters (half the diameter of the base) and the height is 5 meters. Therefore, the volume of the tank is approximately 20.94 cubic meters. The volume flow rate of water exiting the tank can be calculated by multiplying the area of the drain by the velocity of the water: A = (π/4)d² = (π/4)(0.05²m²) ≈ 0.00196 m². Therefore, the volume flow rate is 0.00196 m²/minute * 1 liter/1000 cm³ * 60 seconds/minute ≈ 0.0702 liters per second.
The volume of salt in the solution can be calculated by multiplying the salt concentration by the volume of the solution: V_salt = 0.1 g/ml * 500 liters = 50 grams. The flow rate of salt exiting the tank can be calculated by multiplying the volume flow rate of water by the salt concentration: 1 liter/minute * 0.1 g/ml = 0.1 grams/minute. Therefore, the flow rate of salt is 0.1 grams per minute, or approximately 0.00167 grams per second.
The exit velocity of the water can be calculated using Torricelli’s equation, which relates the exit velocity to the height of the liquid in the tank: v = √(2gh), where v is the exit velocity, g is the acceleration due to gravity (9.81 m/s²), and h is the height of the liquid in the tank measured from the center of the hole. In this case, h = 1.5 meters. Therefore, the exit velocity is v = √(2 * 9.81 m/s² * 1.5 m) ≈ 3.83 m/s. Converted to units of liters per second (L/s), the exit velocity is approximately 38.37 L/s.
As in the previous exercise, the exit velocity of the oil can be calculated with Torricelli’s equation. In this case, h = 4 meters, so the exit velocity is v = √(2 * 9.81 m/s² * 4 m) ≈ 8.85 m/s. Converted to units of liters per second (L/s), the exit velocity is approximately 185.37 L/s.
The speed at which water exits can be calculated using the Bernoulli equation, which relates the water’s velocity at the hole with the liquid pressure on the surface of the tank: v = √(2gh), where v is the exit velocity, g is the acceleration due to gravity (9.81 m/s²), h is the height of the liquid above the hole, and the pressure on the surface of the water is P = ρgh, where ρ is the density of water (1000 kg/m³). In this case, h = 0.8 meters (the height difference between the water and the hole), so the pressure on the surface of the water is P = 1000 kg/m³ * 9.81 m/s² * 0.8 m ≈ 7848 Pa. The exit velocity is then v = √(2 * 7848 Pa / 1000 kg/m³) ≈ 3.53 m/s. Converted to liters per second (L/s), the exit velocity is approximately 17.67 L/s.
Teaching resources
Here are five teaching resources on Torricelli’s theorem along with their links:
Torricelli’s Law – Conceptual Physics by Paul G. Hewitt: This video explains the principles behind Torricelli’s theorem in a simple and intuitive way, making it suitable for high school students. https://www.youtube.com/watch?v=sJ7VjHsGVgU
The Torricelli Theorem: This website provides an interactive demonstration of Torricelli’s theorem, which allows students to visualize the theorem in action. http://www.mathematik.com/Torricelli.html
Torricelli’s Theorem: This website provides a brief overview of Torricelli’s theorem along with examples of its applications, making it useful for students who are interested in learning more about the practical applications of this theorem. https://www.pleacher.com/mp/mlessons/calculus/torricel.html
Let’s imagine that we have a certain amount of mass of a substance at a certain temperature and we mix it with a certain amount of another substance (or the same) at another temperature. What will be the final temperature of the mixture? Are there ways to obtain it? Yes, and the calculation is quite simple to understand if we apply the formulas of calorimetry.
Calorimetry is the branch of thermodynamics (a discipline of physics) that studies the exchanges of heat between a system and others, quantitatively. That is, we can be able to calculate quantities of heat involved in a thermodynamic exchange.
Sensible Heat
o perform calorimetry exercises, we must first know what sensible heat is. To do this, let’s start by defining the term heat.
We know that “Heat is the transfer of energy”, says the definition. The energy that is delivered to a system can have important effects on that system. We have also worked, in previous articles, on other forms of energy transfer such as mechanical work.
Among other effects that we can see daily, we find, for example, that metals expand with heat or that masses of water heat up when they absorb heat. Let’s analyze a little this last mentioned example. The heat that is delivered to the system agitates the water molecules causing their temperature to rise. This energy in transit, which can be easily experienced by measuring the initial and final temperature of a system, is called sensible heat.
It is easy to calculate mathematically, using the formula: Q = m · c · Δt. In the previous formula, Q is the heat involved; m is the mass of the system; c is the called specific heat – which we will explain later -; and Δt is the temperature difference. Let’s remember that Δt can also be written as Δt = tf – ti. In other words, our formula for sensible heat is:
\( Q=m\cdot c\cdot (t_f-t_i) \)
Eq (I)
What is specific heat?
Specific heat is the amount of heat required to raise the temperature of one gram of substance by one degree Celsius. Phew! What does all this mean? Don’t worry, it’s just a unique value for each substance that is tabulated, meaning there are tables (like the one we provide below) where you can look up the values of c for each substance.
Substance
Specific Heat (cal/g.°C)
Oil
0.4
Steel
0.115
Water
1
Saltwater
0.95
Alcohol
0.574
Aluminum
0.226
Ammonia
1.07
Bronze
0.088
Zinc
0.094
Copper
0.094
Tin
0.06
Ether
0.54
Glycerin
0.58
Iron
0.115
Ice
0.489
Brass
0.094
Mercury
0.033
Nickel
0.11
Silver
0.056
Lead
0.035
Petroleum
0.5
Glass
0.2
Knowing these values, we can determine the amount of heat necessary to raise the temperature of a certain mass of a substance. Let’s look at practical examples:
Sensible Heat Solved Exercise
How much heat will be necessary to deliver to 23 grams of iron to raise its temperature from 23°C to 45°C?
It is important to take into account the data offered in the statement. Then, apply the formula provided in equation (I).
In this way, we see that it is necessary to deliver 58.19 cal of energy to 23 g of iron at 23°C to raise its temperature to 45°C.
Latent Heat
Latent heat is the heat necessary to undergo a state change of a certain mass of a substance. In other words, it indicates how much energy must be delivered to a system to change its state of aggregation. In the case that the system changes from a solid state to a liquid state (i.e., melting) or from a liquid state to a solid state (i.e., solidification), we are facing a latent heat of fusion. In the case that the system changes from a liquid state to a gaseous state (i.e., vaporization) or from a gaseous state to a liquid state (i.e., condensation), we are facing a latent heat of vaporization.
The latent heat of fusion is denoted as: \(Q_{L}^{fus}\)
The latent heat of vaporization is denoted as: \(Q_{L}^{vap}\)
Knowing that, the latent heat can be calculated depending on what type of state change our system is undergoing:
In the case of melting or solidification, we must use:
\( Q_{L}^{fus}=m\cdot L_{f}\)
(Equation 2)
…where \(Q_{L}^{fus}\) is the latent heat of fusion; m is the mass of the substance, and \(L_{f}\) is the fusion constant value of the substance involved (it is a unique value for each substance).
In the case of vaporization or condensation, we must use:
\( Q_{L}^{vap}=m\cdot L_{V}\) (Equation 3)
…where \(Q_{L}^{vap}\) is the latent heat of vaporization; m is the mass of the substance, and \(L_{v}\) is the vaporization constant value of the substance involved (it is a unique value for each substance, different from \(L_{f})\).
Latent heat is usually very large because the energy required to break the intermolecular bonds between the molecules of a system is much greater than the energy required to deliver to a system to raise its temperature. Let’s remember that temperature is nothing more than a measure of the kinetic energy of the particles of that system.
Solving Enthalpy Exercises (Heat at constant pressure)
To understand this topic, let’s analyze the example:
25 grams of water are desired to be heated from -23°C to 130°C. How much heat must be supplied? Keep in mind that the specific heat of water is 1 cal/g°C, its fusion heat value is 79.7 cal/g, and its vaporization heat value is 539.4 cal/g.
Firstly, we must take into account that the substance involved here is water, whose specific heat data and fusion and vaporization values are given in the statement. Therefore, let’s write down all the data we have:
Once the data is written, we must analyze the problem.
Temperatura inicial = Initial temperature Temperatura final = Final temperature Temperatura de fusión = Melting temperature Temperatura de ebullición = Boiling temperature
As seen in figure 1, we must list (for better organization) all the temperatures we have, correctly differentiating the initial temperature, the melting temperature, the boiling temperature, and the final temperature, all in the correct order.
From -23°C to 0°C, water increases its temperature in the presence of sensible heat. Then, at 0°C, a change of state occurs (latent heat). From 0°C to 100°C, we have sensible heat again (water increases its temperature). At 100°C, we encounter latent heat because the system is vaporizing. Finally, sensible heat awaits us from 100°C to 130°C.
We will call \(Q_{s}^{1}\) to the first sensible heat (which goes from -23°C to 0°C); \(Q_{s}^{2}\) to the second (which goes from 0°C to 100°C); and \(Q_{s}^{3}\) to the third (which goes from 100°C to 130°C). We will call \(Q_{L}^{1}\) to the first latent heat that appears (at 0°C) and \(Q_{L}^{2}\) to the second (at 100°C).
In conclusion:
Temperatura inicial = Initial temperature Temperatura final = Final temperature Temperatura de fusión = Melting temperature Temperatura de ebullición = Boiling temperature
Now, let’s calculate one by one by replacing the data in equations 1, 2, and 3 as appropriate:
Thanks to what we have learned in this article, we can predict final temperatures of systems that have been supplied with heat.
To better understand everything, let’s first make a theoretical analysis of the case, and then apply it to an example.
A bit of theory about calorimetry.
The formula for sensible heat (that is, the transfer of energy that occurs when a body changes its temperature) is:
\( Q=c_{2}\cdot m_{2}.(T_{f}-T_{i})\)
(Equation 1)
…where Q is the heat gained or lost, c is the specific heat of the substance being studied, m is the mass of the body, Tf is the final temperature of the system, and Ti is the initial temperature of the system.
If two isolated bodies or systems exchange energy in the form of heat, the amount received by one of them is equal to the amount given up by the other body. In other words:
The total exchanged energy is constant; it is conserved.
This means that when we look at heat gains or losses, we immediately find that:
\( \Sigma Q=0\)
In simpler terms, the above equation means that:
\( Q_{2}+Q_{1}=0 \)
If we subtract Q1 from both sides of the equation, we get something quite useful:
\(Q_{2}= -Q_{1} \) (Equation 2)
These values represent the sensible heat of the second and first bodies (\(Q_{2}\) and [/atex]Q_{1}[/latex], respectively).
Two bodies in thermal contact will reach thermal equilibrium after a time (as predicted by the zeroth law of thermodynamics). This means that both bodies will have the same final temperature. Knowing this, we substitute Equation 2 with the factors of Equation 1.
If we add 10 liters of water at 13°C to a 90-liter aquarium with water temperature of 27°C. What temperature remains in the aquarium after adding the water?
We know that the calculations will guide us to the answer.
First, let’s write the data:
m1 = 10,000 g (First, let’s convert liters to grams. As the density of water is 1 g/mL, 1 liter of water is exactly equal to 1 kg of water. But as we said we need the information in grams, then we convert kg to g and ensure we have 10,000 g of water.)
\(c_{1} = 1 cal/g°C\)
\(T_{i_{1}} = 13°C\)
\(m_{2} = 90,000 g \)
\(c_{2} = 1 cal/g°C Ti2 = 27°C \)
Secondly, we apply equation 3, which leads to equation 4 to find the final temperature of thermal equilibrium between the two masses of water:
Here you will find dozens of practical exercises on the topic discussed here, regarding the application of knowledge on sensible and latent heat.
Exercises on Sensible Heat
How much energy (in calories) is necessary to deliver to 1g of water for its temperature to rise from 24°C to 25°C (Hint: cwater=1cal/g°C)? Ans: 1 cal.
Express the previous answer in Joules. Ans: 4.18 J.
How many J correspond to 234 cal? Ans: 978.12 J.
How many cal correspond to 45.6 J? Ans: 10.91 cal.
Is cal the same as Cal? Explain.
Mark T or F: “34000 cal is equal to 34 Cal, so it is also equivalent to 34 Kcal”
How much energy (in calories) is necessary to deliver to 156g of water for its temperature to rise from 14°C to 55°C? Ans: 6396 cal
How much energy (in calories) is necessary to deliver to 123.4g of bronze for its temperature to increase from 45°C to 65.6°C? Given: cbronze=0.086 cal/g°C. Ans: 218.62 cal
How much energy (in calories) is necessary to deliver to 134.5g of oil for its temperature to rise by 34 degrees Celsius (Hint: coil=0.40 cal/g°C)? Ans: 1829.2 cal
How much energy (in calories) is involved in the temperature change from 23°C to 8°C of 45.6 grams of water? Ans: -684 cal
Express the previous answer in kilojoules. Ans: -2.86 kJ
Why is the previous value negative? What does it mean: a) Q<0, b) Q>0, c) Q=0?
How much energy (in calories) is necessary to deliver to 156g of water for its temperature to rise from 14°C to 55°C? Ans: 6396 cal
What will be the final temperature of 34 g of alcohol if its initial temperature was 34°C and 326.4 cal was delivered to the system (Hint: calcohol=0.6 cal/g°C)? Ans: 50°C
Given the following data: m=34g; c=0.2 cal/g°C; Tf=45°C; Ti=34°C, what is the value of Q? Ans: 74.8 cal.
Given the following data: m=54g; c=0.9 cal/g°C; Tf=25°C; Ti=45°C, What is the value of Q? Answer: -972 cal
Given the following data: m=5.6g; c=0.5 cal/g°C; Tf=-25°C; Ti=-30°C, What is the value of Q? Answer: 14 cal
If 1673.4 cal of energy were used to heat a mass of water from 12°C to 14°C, what is the value of that mass of water? Answer: 836.7 g
If 237 cal of energy were used to heat a mass of chromium (ccr=0.108 cal/g°C) from 12°C to 14°C, what is the value of that mass of chromium? Answer: 1097.22 g
234.500g of aluminum were heated by providing 814.184 cal of energy. If cAluminum=0.217 cal/g°C and its initial temperature was 34°C, what is its final temperature? Answer: 50°C
When a mass of 67g of a certain substance (c=1.1 cal/g°C) cooled down from a temperature of -23°C, a value of Q=-147.4 cal was obtained. What is the final temperature of the system? Answer: -25°C.
What is the value of c of a substance if 458 cal of energy were required to go from 54.3°C to 67.3°C for a sample of 202g of that substance? Answer: 0.17 cal/g°C
Given the following data: m=5.6g; Q=34.5 cal; Ti=-25°C; Tf=-20°C, What is the value of c? Answer: 1.23 cal/g°C
Calorimetry allows the calculation of quantities of heat involved in thermodynamic processes.
Exercises in Calorimetry with Latent Heat
What is latent heat and how does it differ from sensible heat? Explain in detail why the amount of heat that needs to be supplied to a system to change its state of aggregation is usually very large.
What do we call regressive and progressive changes in state of aggregation?
Correctly name all the changes of state (regressive and progressive) that occur between a solid, a liquid, and a gas.
Knowing that the heat of fusion of a substance is 34.5 cal/g, how much energy will need to be supplied to 23 g of that substance to change from a solid to a liquid state? Ans: 793.5 cal
Knowing that the heat of vaporization of a substance is 342.5 cal/g, how much energy will need to be supplied to 453 g of that substance to change from a liquid to a gaseous state? Ans: 155,152.5 cal
How much energy will need to be supplied to 45.6 g of water to change from a liquid to a gaseous state, knowing that Lvap=539.4 cal/g and Lfus=79.7 cal/g? Ans: 24,596.64 cal
How much energy is involved in the change of state from gas to liquid of 45 g of ammonia whose values of L are: Lvap=327 cal/g and Lfus=180 cal/g? Ans: 14,715 cal
What is the value of Lvap for a substance for which 5673 cal of energy is needed to change 64 g of that substance from the liquid to the gaseous state? Ans: 88.64 cal
Exercises of Calorimetry involving changes of state.
How much energy is needed to be delivered to 23g of water to go from 84°C to 120°C, knowing that Lvap=539.4 cal/g and the boiling point of H2O is 100°C? Ans: 13,234.2 cal
How much energy is needed to be delivered to 46g of water to go from -4°C to 10°C, knowing that Lfus=79.7 cal/g and the melting point of H2O is 0°C? Ans: 4,310.2 cal
Given the following data for a substance: m=34g; c=0.3 cal/g°C; Lfus=345.5 cal/g; Tfus=34°C; Ti=30°C; Tf=45°C, what is the value of Q, that is, the energy needed for the temperature change from the initial to the final state? Ans: 11,900 cal
Given the following data for a substance: m=5.4g; c=0.6 cal/g°C; Lfus=345.5 cal/g; Lvap=245.4 cal/g; Tfus=-34°C; Tebul=344°C; Ti=300°C; Tf=363°C, what is the value of Q? Ans: 1,529.28 cal
Knowing c, Tebul, Tfus, Lvap, and Lfus of water given in the previous exercises, how much energy is needed to be delivered to 34g of H2O to go from -30°C to 134°C? Ans: 26,625.4 cal
Given the following data for a substance: m=5.4g; c=0.6 cal/g°C; Lfus=345.5 cal/g; Lvap=245.4 cal/g; Tfus=-34°C; Tebul=4°C; Ti=-40°C; Tf=10°C, what is the value of Q? Ans: 3352.86 cal
Given the following data for a substance: Q=3456 cal; c=0.6 cal/g°C; m=9.6 g; Lvap=245.4 cal/g; Tfus=-3.4°C; Tebul=34.4°C; Ti=-4°C; Tf=36.3°C, what is the value of Lfus? Ans: 90.42 cal/g
Recommended Mesography
The website Educaplus.org presents a very interesting applet on the subject, available at http://www.educaplus.org/game/calorimetria. In this application, you can verify what happens to the temperature of water when masses at different temperatures are added to it. You can even check the results of problems.
These are frequently asked questions in the world of Physical Education. Let’s take a look at them!
1. What are the physical qualities or abilities?
The physical qualities or abilities are: strength, speed, endurance, elongation, and/or stretching.
2. What is meant by endurance in Physical Education?
Endurance is the ability to make an effort of greater or lesser intensity for a prolonged period, as well as the ability to resist fatigue (whether biological, cerebral, anatomical, etc.) that an individual has.
3. What types of endurance exist in Physical Education?
The types of endurance are:
Aerobic Endurance (which is defined as the ability to maintain continuous effort for a long time, during which oxygen supply to the blood allows for the muscular expenditure needs to be met).
Anaerobic Endurance (which is defined as the ability to perform intense exercise, causing an imbalance between the supply of oxygen and the body’s needs). Within this last type, two subtypes are distinguished: Alactic Anaerobic Endurance, in which ATP and P. C. (energy) stored in the muscle are used without oxygen debt and without waste production; and Lactic Anaerobic Endurance, in which lactic acid formation occurs because the breakdown of sugars and fats to resynthesize ATP through glycolysis (the process of breaking down glucose into its simplest components within cells) is carried out with oxygen debt.
Fatigue is an alarm reaction that manifests itself in the individual, causing a decrease in performance due to an excess of stimulation.
5. What are the types of fatigue?
The types of fatigue are three: psychological fatigue, nervous fatigue, and muscular fatigue.
6. What is meant by muscle quality?
Muscle quality refers to the fact that the muscle has larger or smaller motor units, its fibers are of a certain type, etc. It depends on the training being better assimilated and the endurance performances being higher.
7. What are the energy systems?
The energy systems are: maximal intensity efforts (Alactic Anaerobic Endurance), submaximal intensity efforts (Lactic Anaerobic Endurance), and medium intensity efforts (Aerobic Endurance).
8. What is the heart rate for each one?
The heart rate for each one is: in maximal intensity efforts, above 180 beats per minute; in submaximal intensity efforts, above 140 beats per minute, and in medium intensity efforts, between 120 and 140 beats per minute.
Physical Education. Sports are essential for balanced health.
9. According to the physio-biological characteristics, what are the phases from 13 to 17 years old?
According to the physio-biological characteristics, there are two phases from 13 to 17 years old, which are differentiated as follows: one lasts until 14 years old (puberty produces a minimal capacity to withstand effort), and the other goes from 14 to 17 (in which the work of aerobic endurance increases and anaerobic endurance begins).
10. What are the different types of effort?
Regarding the different types of efforts, we can classify them into the alactic system (speed exercises up to 7 seconds, pure speed characterized as “power,” and from 7 to 15 seconds, prolonged speed known as “capacity”), which is anaerobic but without lactate production -the recovery time needed is about 2 to 3 minutes, depending on each individual’s preparation-; the lactate system involves the segregation of lactic acid, which causes cramps, from exercises lasting more than 15 seconds.
Examples of this would be, in the first case, speed exercises such as running a certain distance (whether it’s 50 meters or other) in a time less than 15 seconds. The second case is practically the same, but with an even greater distance in a time that exceeds 15 seconds.
You can find more content on gamification in the classroom in Related Content.
Drosophila melanogaster, also known as the fruit fly, has been a valuable tool in scientific research for over a century. Its short lifespan, easy handling, and ability to produce large quantities of offspring have allowed scientists to study a variety of biological processes, including genetics, development, physiology, and behavior. In addition, the genome of Drosophila has been fully sequenced, which has allowed for a better understanding of molecular biology and the relationship between genes and human diseases. Drosophila melanogaster continues to be a valuable tool for scientific research and has been instrumental in advancing our understanding of biology and genetics.
Modeling of Drosophila melanogaster. Created by AI.
Why use Drosophila Melanogaster (or fruit fly) in genetic experiments?
D. melanogaster is an ideal organism for experimentation for several reasons. Firstly, it is abundant and easy to capture, and can be easily cultivated in the laboratory. Additionally, it produces a large number of offspring, making it suitable for testing Mendelian ratios. Its biological cycle is completed in 10-11 days at 25°C, and it has only 4 pairs of chromosomes, 3 autosomes and 1 sexual, with sex determination of XX females and XY males. Since its use in 1905, an abundant bibliography has been accumulated. Finally, there are a large number of natural and induced mutants and special strains that allow for careful genetic analysis.
What are the mutations present in Drosophila Melanogaster?
There are several known mutations in Drosophila melanogaster. These are the most common mutations and their characteristics:
“White-eye” mutation: This mutation is due to a recessive gene on the X chromosome that causes the absence of pigment in the eyes, resulting in white eyes instead of red.
“Sepia” mutation: This mutation also affects eye color, but instead of resulting in white eyes, it produces dark brown eyes.
“Vestigial” mutation (short wing): This mutation affects wing development and produces shorter and more wrinkled wings than normal.
“Bar” mutation: This mutation affects the development of hairs on the fly’s legs and produces shorter and thicker hairs.
“Antennapedia” mutation: This mutation affects antenna development and produces a transformation of the legs into antennas.
“Flamingo” mutation: This mutation produces defects in the formation of epithelial cells, resulting in deformed legs.
“Ubx” mutation: This mutation affects leg development and produces additional or fused legs in different parts of the body.
These are just some of the known mutations in Drosophila melanogaster. There are many other mutations that affect the shape, color, and behavior of fruit flies. Mutations in Drosophila melanogaster have been an important tool in genetic research for decades.
Eye mutations in Drosophila melanogaster.
Modeling of Drosophila melanogaster. Created by AI.
Drosophila melanogaster is a species commonly used in genetic studies due to its short life cycle and ability to produce large amounts of offspring. In this species, there are several known mutations that affect the appearance of the eyes. Below is a table with some of these mutations and their characteristics:
Mutation
Description
White eyes (w)
The eyes are completely white and lack pigmentation. This mutation is due to the lack of an enzyme necessary to produce pigment in the eyes.
Red eyes (w+)
The eyes are dark red in color and have normal pigmentation. This is the wild type or normal form of the gene that produces the enzyme necessary to produce pigment in the eyes.
Scarlet eyes (st)
The eyes are dark red in color and have a mottled or speckled appearance. This mutation is due to the lack of a protein normally found in the pigment cells of the eyes. The unpigmented areas are due to cell death.
Ruby eyes (ru)
The eyes are dark red in color and have a shiny, translucent appearance. This mutation is due to a lack of pigment and an alteration in eye structure. Ruby eyes have a unique appearance and are often used in studies of eye development.
Giant eyes (g)
The eyes are abnormally large and have a bulging appearance. This mutation is due to the excessive production of a protein that controls eye growth.
Small eyes (ey)
The eyes are abnormally small and have a shriveled appearance. This mutation is due to the lack of a protein normally found in the pigment cells of the eyes. The lack of this protein causes the eye cells not to divide properly, resulting in smaller eyes.
It is important to note that there are many more mutations that affect the appearance of the eyes in Drosophila melanogaster, and each mutation can have additional effects on the health and behavior of the fly. Additionally, the expression of these mutations may be influenced by additional environmental and genetic factors.
Procedure
Collect flies in each house and place them in glass jars containing food (decaying fruit). Cover with cotton or gauze and a rubber band.
Bring the jars with the flies to school (always have reserve jars at home). Search for information on fly characteristics on the Internet, print it in color or draw it.
Look for information on mutations. Submit partial report.
Put the flies to sleep and start noting their characteristics, as well as separating males from females. Note eye color and wing size characteristics.
Prepare jars for pairs: at least 5 jars per group.
After sterilizing the jar with alcohol, put the food and then introduce the pair of flies. Label the jar with the characteristics, date of assembly, and group name. If changing the food, notify the teacher and/or laboratory assistant.
Store the jar in a warm place. Observe every two days. When the eggs appear, remove the pair and wait for the births.
Observe the larvae, look at them with a magnifying glass and draw them. Look for information on the larvae on the Internet. Submit partial report.
Observe the pupae, describe them and note how many days they are in that stage.
Observe the born flies, count and draw one. Separate them into male and female. Count them according to eye color and wing size.
Note if there are mutants and separate them. The non-mutants should be reserved separately and classified as “Generation F1”.
Extract a new pair from this F1 to obtain the F2 and proceed in the same way in each case.
Always consider hygiene, sterilization, food preparation, care when handling, and order in the group for the procedure.
Always take notes on observations, stay informed about the zoology of the flies, and apply what was learned in theoretical classes.
Conclusion questions:
What were the characteristics observed in fruit flies in this experiment and how were these characteristics transmitted from one generation to another?
How can a fruit fly’s genotype be determined based on its phenotypic characteristics?
What is the importance of fruit flies in scientific research and how have they been used in genetics and other fields of biology?
What role do genes play in determining physical and behavioral characteristics in fruit flies and how can these characteristics affect the survival and reproductive success of the flies?
What are some of the limitations and challenges associated with the use of fruit flies in scientific research and how can they be overcome.
Sources
Here are some sources where it is possible to find simulations or videos about the importance of Drosophila melanogaster in genetics:
Drosophila genetics tutorial: This tutorial is a step-by-step guide to the genetics of Drosophila melanogaster, with explanations and interactive activities. You can find it at: https://www.genetics.org/content/147/3/1189.full
The Virtual Fly Lab: This website provides a range of interactive simulations and activities related to Drosophila genetics, including breeding experiments, mapping genes, and observing mutant phenotypes. You can find it at: http://virtualflylab.org/
The importance of Drosophila in genetics: This video from the National Human Genome Research Institute explains why Drosophila melanogaster is an important model organism for genetic research, and how its use has contributed to our understanding of genetics. You can find it at: https://www.youtube.com/watch?v=0gZ5oGQ5I5g
Fruit Fly Genetics – Drosophila melanogaster: This video from the Dolan DNA Learning Center provides a clear explanation of the use of Drosophila melanogaster in genetics research, and how it has led to important discoveries. You can find it at: https://www.youtube.com/watch?v=v1lr9n1pMvE
La Drosophila melanogaster, también conocida como mosca de la fruta, ha sido una herramienta valiosa en la investigación científica durante más de un siglo. Su corto ciclo de vida, su fácil manejo y su capacidad de producir grandes cantidades de descendencia han permitido a los científicos estudiar una variedad de procesos biológicos, incluyendo la genética, el desarrollo, la fisiología y el comportamiento. Además, el genoma de la Drosophila ha sido completamente secuenciado, lo que ha permitido una mejor comprensión de la biología molecular y la relación entre los genes y las enfermedades humanas. La Drosophila melanogaster sigue siendo una herramienta valiosa para la investigación científica y ha sido fundamental en el avance de nuestra comprensión de la biología y la genética.
Modelización de Drosophila melanogaster. Creado por IA.
¿Por qué usar Drosophila Melanogaster (o mosca de la fruta) en experiencias de genética?
D. melanogaster es un organismo ideal para la experimentación por varias razones. En primer lugar, es abundante y fácil de capturar, y puede ser cultivado fácilmente en el laboratorio. Además, produce una gran cantidad de descendientes, lo que lo hace adecuado para comprobar las proporciones mendelianas. El ciclo biológico se completa en 10-11 días a 25°C, y posee solo 4 pares de cromosomas, 3 autosomas y 1 sexual, con determinación del sexo XX hembras y XY machos. Desde su uso en 1905, se ha acumulado una abundante bibliografía. Finalmente, hay una gran cantidad de mutantes naturales e inducidos y cepas especiales que permiten análisis genéticos cuidadosos.
¿Cuáles son las mutaciones que presenta la Drosophila Melanogaster?
Existen varias mutaciones conocidas en Drosophila melanogaster. Estas son las mutaciones más comunes y sus características:
Mutación “white-eye” (ojo blanco): Esta mutación se debe a un gen recesivo en el cromosoma X que causa la ausencia de pigmento en los ojos, lo que resulta en ojos blancos en lugar de rojos.
Mutación “sepia“: Esta mutación también afecta el color de los ojos, pero en lugar de resultar en ojos blancos, produce ojos marrones oscuros.
Mutación “vestigial” (ala corta): Esta mutación afecta el desarrollo de las alas y produce alas más cortas y arrugadas que las normales.
Mutación “bar“: Esta mutación afecta el desarrollo de los pelos en las patas de la mosca y produce pelos más cortos y gruesos.
Mutación “antennapedia“: Esta mutación afecta el desarrollo de las antenas y produce una transformación de las patas en antenas.
Mutación “flamingo“: Esta mutación produce defectos en la formación de las células epiteliales, resultando en patas deformes.
Mutación “Ubx“: Esta mutación afecta el desarrollo de las patas y produce patas adicionales o fusionadas en diferentes partes del cuerpo.
Estas son solo algunas de las mutaciones conocidas en Drosophila melanogaster. Hay muchas otras mutaciones que afectan la forma, el color y el comportamiento de las moscas de la fruta. Las mutaciones en Drosophila melanogaster han sido una herramienta importante en la investigación genética durante décadas.
Mutaciones de ojos en las Drosophila melanogaster
Modelización de Drosophila melanogaster. Creado por IA.
La Drosophila melanogaster es una especie comúnmente utilizada en estudios genéticos debido a su corto ciclo de vida y su capacidad para producir grandes cantidades de descendencia. En esta especie, hay varias mutaciones conocidas que afectan la apariencia de los ojos. A continuación, se presenta un cuadro con algunas de estas mutaciones y sus características:
Mutación
Características
Ojos Blancos (w)
Los ojos son completamente blancos y carecen de pigmentación. Esta mutación se debe a la falta de una enzima necesaria para producir el pigmento en los ojos.
Ojos Rojos (w+)
Los ojos son de color rojo oscuro y tienen una pigmentación normal. Esta es la forma salvaje o normal del gen que produce la enzima necesaria para producir el pigmento en los ojos.
Ojos Escarlatas (st)
Los ojos son de color rojo oscuro y tienen una apariencia moteada o moteada. Esta mutación se debe a la falta de una proteína que normalmente se encuentra en las células pigmentarias de los ojos. Las áreas sin pigmento se deben a la muerte celular.
Ojos de Rubí (ru)
Los ojos son de color rojo oscuro y tienen una apariencia brillante y translúcida. Esta mutación se debe a una falta de pigmento y a una alteración en la estructura del ojo. Los ojos de rubí tienen una apariencia única y se utilizan a menudo en estudios de desarrollo de los ojos.
Ojos Gigantes (g)
Los ojos son anormalmente grandes y tienen una apariencia abultada. Esta mutación se debe a la producción excesiva de una proteína que controla el crecimiento de los ojos.
Ojos Pequeños (ey)
Los ojos son anormalmente pequeños y tienen una apariencia encogida. Esta mutación se debe a la falta de una proteína que normalmente se encuentra en las células pigmentarias de los ojos. La falta de esta proteína hace que las células de los ojos no se dividan adecuadamente, lo que resulta en ojos más pequeños.
Es importante tener en cuenta que hay muchas más mutaciones que afectan la apariencia de los ojos en la Drosophila melanogaster, y cada mutación puede tener efectos adicionales en la salud y el comportamiento de la mosca. Además, la expresión de estas mutaciones puede estar influenciada por factores ambientales y genéticos adicionales.
Procedimiento
Recolectar moscas en cada casa y colocar en frascos de vidrio que contengan alimento (fruta en estado de descomposición). Tapar con algodón o gasas y banda elástica.
Traer a la escuela los frascos con moscas (siempre tener en sus casas frascos de reserva).
Buscar información sobre las características de las moscas en Internet, imprimirla a color o dibujarlas. Buscar información sobre las mutaciones. Entregar informe parcial.
Dormir las moscas y empezar a anotar sus características, además de separar machos de hembras. Anotar características de color de ojos y tamaño de alas.
Preparar los frascos para las parejas: mínimo 5 frascos por grupos.
Colocar el alimento después de haber esterilizado el frasco con alcohol y luego introducir a la pareja de moscas, rotular el frasco anotando las características, fecha de armado y nombre del grupo. Si cambia de alimento, avisar al docente y/o auxilar de laboratorio.
Guardar el frasco en lugar templado.
Observar cada dos días.
Cuando aparecen los huevos, sacar la pareja y esperar los nacimientos.
Observar las larvas, mirarlas con lupa y dibujar. Buscar información en Internet sobre las mismas. Entregar informe parcial.
Observar las pupas, describirlas y anotar cuántos días están.
Observar las moscas nacidas, contarlas y dibujar alguna. Separarlas en macho y hembra. Contarlas según el color de ojos y tamaño de alas.
Anotar si hay mutantes y separarlas. Las que no son mutantes, reservarlas aparte y clasificarlas como “Generación F1”.
Extraer de esta F1 una nueva pareja para obtener la F2 y proceder de la misma manera en cada caso.
Tener en cuenta siempre para el procedimiento, todos los cuidados de higiene, esterilización, preparación de alimentos, cuidado al manipularlas, orden en el grupo.
Siempre llevar nota de lo observado, mantenerse informado sobre la zoología de las moscas y aplicar lo aprendido en las clases teóricas.
Preguntas de conclusión:
¿Cuáles fueron las características observadas en las moscas de la fruta en este experimento y cómo se transmitieron estas características de una generación a otra?
¿Cómo se puede determinar el genotipo de una mosca de la fruta en base a sus características fenotípicas?
¿Cuál es la importancia de las moscas de la fruta en la investigación científica y cómo se han utilizado en la investigación de la genética y otros campos de la biología?
¿Qué papel juegan los genes en la determinación de las características físicas y de comportamiento en las moscas de la fruta y cómo pueden afectar estas características la supervivencia y el éxito reproductivo de las moscas?
¿Cuáles son algunas de las limitaciones y desafíos asociados con el uso de moscas de la fruta en la investigación científica y cómo pueden ser superados?
Más información:
Simulación interactiva sobre la genética de Drosophila melanogaster: este recurso en línea proporcionado por la Universidad de Utah permite a los usuarios interactuar con una simulación de cruzamiento de Drosophila melanogaster para comprender mejor los principios genéticos clave. Disponible en: https://learn.genetics.utah.edu/content/labs/fly/
Canal de YouTube “BioEnLaRed”: este canal de YouTube en español cuenta con varios videos sobre la genética de Drosophila melanogaster, incluyendo explicaciones de los experimentos de Thomas Hunt Morgan y cómo se utilizan las moscas de la fruta en la investigación científica. Disponible en: https://www.youtube.com/channel/UCkCE1BiB0yxnF_UjGvXpnHA
Artículo sobre Drosophila melanogaster en el sitio web “Biología-Geología”: este artículo en español proporciona una descripción general de la importancia de Drosophila melanogaster en la investigación científica, incluyendo su uso en la genética y otros campos de la biología. Disponible en: https://www.biologia-geologia.com/biologia/drosophila-melanogaster
This article about Scientific Traditions in the Renaissance aims to explain the attempts by Renaissance figures to “order the disorder” and “find light amidst the scientific entropy” caused by our lack of understanding of the world around us. During this fruitful period of knowledge, three worldviews or scientific traditions coexisted: organicism, neoplatonism, and mechanicism. Let’s take a closer look at each of them in this article on the history of science!
The three worldviews or scientific traditions
Organicism
In organicism, the specific qualities of things and living beings, that is, beings with life, allow the knowledge of reality. In other words, what allows us to know nature is merely perceived by the senses. It is evident that this leads us to problems and multiple errors. If current scientific knowledge followed the laws of organicism, erroneous theories like the heliocentric one would continue to explain our place in the universe, since our eyes perceive how the Sun moves through the skies as if it revolved around us, on Earth.
In this sense, the investigation of nature will not depend on abstractions from fields such as mathematics. In fact, we could say that mathematics does not play a fundamental role in explaining natural phenomena since it cannot replace immediate experience. This vision, as students should be told -especially after having brought to mind what was learned about the Renaissance-, assumes the continuation of medieval scholastic thought.
Once the characteristics of organicism are understood, we will be able to explain to students how this changed in a short time: mathematics then becomes more than important. Neoplatonism and mechanism argue that the secrets of nature are written in the language of mathematics.
Neoplatonism
With references to ancient and pagan schools such as Pythagorean and Hermetic, neoplatonism argued that whoever wanted to explain the great natural secrets should act like a magician. The neoplatonics accepted the ideas of Nicolaus Copernicus (1473-1543) and proclaimed that the source of knowledge was mystical contemplation of the world.
Some representatives of neoplatonism were Giordano Bruno (1548-1600) and Johannes Kepler (1571-1630). As we know, Kepler formulated important laws of the motion of the stars, thus bringing astronomy to a mystically relevant place.
Johannes Kepler: one of the most important exponents of neoplatonism.
Mechanism
Finally, mechanism argued that the universe functioned like a “great machine” that could be analyzed according to its parts and thus understand its functioning. Mathematics was undoubtedly the language in which nature itself was written, but it did not have a mystical sense, as neoplatonism did. The productive needs of the time were the target of the workshops of the craftsmen. It was precisely in these workshops where mechanics, one of the most important branches of classical physics, was born.
Measurement procedures had to be as precise as possible. On the other hand, quantitative relationships had to be established between the measurements obtained.
Galileo Galilei (1542-1642)
This is where Galileo Galilei (1564-1642) comes in: he was the first to introduce the experimental mathematical method in physics. His mechanistic tradition predominated over other traditions. The practical purposes of modern science and its methodological aspects had their foundations in the works of René Descartes and Francis Bacon. Likewise, Leonardo da Vinci, a high-caliber engineer and artist of history, helped to displace Aristotelian philosophy from the place it occupied for centuries.
Activities Proposal for Teachers
Activities
In this project, students will discover Galileo’s contributions to the world of physics. His interests in researching topics such as the fall of bodies and topography, his most notable inventions such as the machine for lifting water and the thermoscope, his research, observations, and the historical context in which he was immersed. It is in his invention of the thermoscope where his work will be connected to topics such as Thermodynamics, where contents related to heat and temperature are studied.
Museum of Arts and Measures. Galileo’s thermoscope (1592)
The students will be divided into groups. They will have to create posters that effectively and attractively inform about Galileo Galilei’s contributions. In these posters, they must present the three philosophies discussed and explained in previous group and oral debates in class, as well as the importance of Galilei in the world of natural sciences.
The evaluation of content will be carried out not only through the timely submission of the requested activities, but also through directed dialogue, oral debate, and sharing of content studied in class.
Discover the differences between CGS, MKS, and Technical Measurement Systems. This article explores the three main systems. Each system has its own advantages and applications, so it’s essential to know which one to use for a specific project. By the end of this article, you’ll have a clear understanding of the different measurement systems and how they impact your work.
In physics and other disciplines of natural sciences, it is essential to have values and units to describe the different magnitudes of your daily life. For example, it is important to mention that it is 34°C on a hot day. If it were 34°F, everything would be completely different: it’s equivalent to 1.1°C! This undoubtedly indicates a very, very cold time. How strange, isn’t it? That’s how important units are!
It is also easy to see that units are related to each other. For example, mass and weight are related to each other, but each of these variables can be measured in different units that, in turn, are related to each other (what a tongue twister!). In this article, we will try to decipher what all this means that we have said.
To establish a system of units, three fundamental magnitudes are required, which can then be related to each other. We begin by saying that physicists have chosen to work with three different measurement systems of units, based on which magnitudes each system chooses as fundamental. Let’s take a look at each of them:
MKS, CGS, and Technical (or Engineering) systems
MKS system
Also called the absolute system, the MKS is a measurement system based on units of LENGTH, MASS, and TIME.
The acronym MKS refers precisely to the words meter, kilogram, and second, which are the base units of this subsystem. Lengths are measured in meters, mass is measured in kilograms, and time is measured in seconds. This is quite useful for measuring magnitudes whose dimensions are often large, such as the width of a house, the height of a giraffe, or the mass of a whale.
Here is a table with the different magnitudes and their units, from the point of view of the MKS system:
Magnitude
Units in MKS
Distance (d)
Meter (m)
Time (t)
Second (s)
Velocity (v)
Meter per second (m/s)
Acceleration (a)
Meter per second squared (m/s²)
Mass (m)
Kilogram (kg)
Force (f)
Newton (N)
CGS system
Also known as the centimeter-gram-second system, the CGS is a measurement system is the other of the measurement systems based on LENGTH, MASS, and TIME. However, the abbreviation CGS refers to the words centimeter, gram, and second. Which of the three units is shared with the MKS system?
Lengths are measured in centimeters, mass is measured in grams, and time, in seconds. This subsystem is not widely used, as it is only useful for measuring magnitudes whose dimensions are small, such as the width of your cell phone, the mass of a worm, or the energy of a butterfly’s flutter.
Here is a table with the different magnitudes and their units, from the point of view of the MKS system:
Magnitude
Unit
Distance (d)
Centimeter (cm)
Time (t)
Second (s)
Velocity (v)
Centimeter/Second (cm/s)
Acceleration (a)
Centimeter/Second^2 (cm/s^2)
Mass (m)
Gram (g)
Force (f)
Dyne (dyn)
Sistema TÉCNICO
The technical system is another measurement system and the last one we will see in this article. It is also called the gravitational system and differs radically from the previous two in that its fundamental magnitudes are no longer length, mass, and time, but rather LENGTH, WEIGHT, and TIME. We have highlighted in italics the magnitude that sets the technical system apart from the others.
Its fundamental units are meter, kilogram-force, and second. Ignoring the fact that it uses kilogram-force as a fundamental unit, which system does it resemble more: MKS or CGS?
What is a kilogram-force?
Here is a table with the different magnitudes and their units, from the point of view of the technical system.
Magnitude
Unit
Distance (d)
Meter (m)
Time (t)
Second (s)
Velocity (v)
Meter/Second (m/s)
Acceleration (a)
Meter/Second^2 (m/s^2)
Mass (m)
Kilogram-force (kgf)
Force (f)
Kilogram-meter/Second^2 (kgm/s^2)
Resources
Heinemann, A. G. (1988). Física: Mecánica-Fluidos-Calor (No. 53). Ángel Estrada.
Here are some resources where you can find simulations and videos about measurement systems and units:
PhET Interactive Simulations – This website offers a range of interactive simulations on various science and engineering topics, including measurement systems and units. You can explore simulations on topics such as converting units, measuring distance, and more. Available on https://phet.colorado.edu/en/simulations/category/physics/measurement
National Institute of Standards and Technology (NIST) – NIST is a United States government agency that is responsible for promoting measurement science and standards. Their website provides access to a range of resources related to measurement systems and units, including videos, tutorials, and interactive tools. Available on https://www.nist.gov/topics/physics/measurement-and-units
The Physics Classroom – The Physics Classroom is an online resource for high school physics students and teachers. They have a section on measurement and units, which includes interactive simulations, tutorials, and practice problems. Available on https://www.physicsclassroom.com/class/measurement
NASA Education – NASA offers a range of educational resources for students and teachers, including videos and interactive simulations on various topics related to space and science. They have a section on measurement and units, which includes simulations on measuring distance and time, and converting units. Available on https://www.nasa.gov/audience/foreducators/topnav/materials/listbytype/Measurement_Resources.html
These resources should provide you with a good starting point for exploring simulations and videos related to measurement systems and units.
Ribonucleic acid (RNA) is an essential molecule for life that is present in all living organisms. This molecule is similar to DNA in its chemical structure, but has some unique characteristics that make it very important for the functioning of cells.
RNA is composed of a chain of nucleotides that contain a nitrogenous base, a sugar, and a phosphate group. The nitrogenous bases are adenine, guanine, cytosine, and uracil, and the sequence of these bases determines the genetic information contained in the RNA.
RNA is synthesized from a DNA strand during the process of transcription and is then used as a template for protein synthesis during the process of translation. Additionally, RNA also has regulatory and structural functions within the cell.
Modelos de ADN y ARN.
There are various types of RNA involved in protein biosynthesis in cells. Messenger RNA (mRNA) carries genetic information from DNA to ribosomes, where it is translated into protein. Transfer RNA (tRNA) brings amino acids to the ribosome, where they are assembled into a polypeptide chain according to the instructions carried by the mRNA. Ribosomal RNA (rRNA) is a component of the ribosome itself and helps catalyze the formation of peptide bonds between amino acids.
Types of RNA
Messenger RNA (mRNA)
The mRNA i is a long single-stranded molecule that contains the codons (a triplet of nucleotide bases) that will be translated into a sequence of amino acids for a protein. Its synthesis takes place in the nucleus and it enters the cytoplasm through the pores of the nuclear envelope. In the cytoplasm, messenger RNA (mRNA) binds to ribosomes, where its codons are translated into the language of amino acids contained in proteins. (One can imagine mRNA as a “molecular photocopy” of the DNA gene.)
Ribosomal RNA (rRNA)
Ribosomes are composed of ribosomal RNA and a large number of proteins. Each ribosome is composed of two subunits: a larger one and a smaller one. The smaller subunit recognizes and binds to messenger RNA and a portion of transfer RNA. The larger subunit consists of three molecules of messenger RNA and several proteins. It contains an enzymatic region that catalyzes the addition of amino acids to the growing protein chain and two catalytic sites (designated P and A) that bind to transfer RNA.
The function of ribosomal RNA is to recognize messenger RNA and catalyze the formation of peptide bonds between the amino acids of the protein.
Transfer RNA (tRNA)
Transfer RNA molecules bind amino acids and deliver them to the ribosome, where they are incorporated into protein chains. There are many types of transfer RNA, at least one type for each amino acid. Their function is to decipher the codons of messenger RNA and translate them into the amino acids of proteins. Transfer RNA molecules have a clover-like structure with a stem.
The outer part of the central leaf contains three exposed bases, called an anticodon, which decipher the code of messenger RNA and is complementary to the codon of messenger RNA that specifies the amino acid to which the transfer RNA is bound. For example, the GUA codon of messenger RNA is complementary to the CAU anticodon of a transfer RNA molecule that carries the amino acid Valine. All of these processes require energy that is provided by the ATP molecule.
Protein synthesis
Stages of protein synthesis.
Protein synthesis occurs in two stages. First, during transcription, messenger RNA is transcribed from the DNA template of the genes in the nucleus. The mRNA travels to a ribosome in the cytoplasm. Second, during translation, the ribosome attaches to the mRNA and the appropriate transfer RNAs.
protein biosynthesis is a fundamental process in the cell that is carried out by translating the genetic information contained in DNA into the language of amino acids. The process of protein synthesis consists of three main stages: initiation, elongation, and termination. The process is explained step by step below:
DNA Transcription: First, DNA is transcribed into messenger RNA (mRNA). This process involves the synthesis of an mRNA molecule that is complementary to one of the two DNA strands. The mRNA contains genetic information that will be used for protein biosynthesis. The genetic code is the set of rules that determine how genetic information is translated into sequences of amino acids. In this code, each group of three nucleotides (known as a codon) codes for a specific amino acid. For example, the codon AUG codes for the amino acid methionine.
Initiation: The process of protein synthesis begins with initiation. The ribosome binds to the 5′ end of the mRNA, recognizing the AUG initiation codon. A tRNA loaded with the amino acid methionine binds to the initiation codon.
Elongation: The next stage is elongation, in which amino acids are added to the growing chain. The ribosome moves along the mRNA, reading each codon and bringing the corresponding tRNA. The amino acid carried by the tRNA is added to the growing chain through a peptide bond. The ribosome iteratively moves along the mRNA, adding one amino acid after another, until it reaches a stop codon.
Termination: When the ribosome reaches a stop codon, protein synthesis ends. At this point, the chain of amino acids is released from the ribosome and folds into a functional protein.
In summary, RNA is a key molecule in cell biology, with essential functions in protein synthesis and gene regulation. Its unique characteristics, such as the presence of uracil instead of thymine and the ability to fold into complex three-dimensional structures, make it a very versatile and adaptable molecule to the needs of the cell.
Synthesis of proteins is a complex process that involves the translation of genetic information from mRNA into the language of amino acids, following the genetic code. Initiation, elongation, and termination are the three main stages of the process. During elongation, the ribosome reads each codon of the mRNA and adds the corresponding amino acid to the growing chain. When the ribosome reaches a stop codon, protein synthesis ends and the chain of amino acids is released to form a functional protein.
The genetic code
The Genetic Code.
Example of protein biosynthesis from a given gene.
Given the following nucleotide sequence in DNA:
TAC CGG AAA CTT AGG GCT ACA CTG CTA TTA
…we will carry out protein biosynthesis.
To obtain the complementary RNA sequence to a DNA sequence, it must be taken into account that the nitrogenous bases adenine (A) pairs with uracil (U) and cytosine (C) pairs with guanine (G).
So, the complementary RNA sequence to the given DNA sequence would be:
AUG GCC UUU GAA UCC CGA UGU GAC GAU UAA
This gene contains 10 codons of three nucleotides each, which are the basic units that the ribosome reads to synthesize the corresponding protein. To simplify the example, let’s assume that this gene codes for a 10 amino acid protein.
The transcription of this gene would result in a complementary mRNA molecule of 30 nucleotides that would bind to the ribosome to begin protein synthesis. Next, a polypeptide chain of 10 amino acids would be assembled from the mRNA codons.
The translation of this gene would follow the following sequence of codons and amino acids:
Codon
Amino Acid
AUG
Methionine
GCC
Alanine
UUU
Phenylalanine
GAA
Glutamic acid
UCC
Serine
CGA
Arginine
UGU
Cysteine
GAC
Aspartic acid
GAU
Aspartic acid
UAA
Stop codon
As can be seen, the sequence of codons in the mRNA dictates the sequence of amino acids in the resulting protein. In this case, the protein contains Methionine, Alanine, Phenylalanine, Glutamic acid, Serine, Arginine, Cysteine, Aspartic acid, Aspartic acid, and a stop codon. Once protein biosynthesis is complete, the polypeptide chain would be released from the ribosome and folded to form a functional three-dimensional protein.
Activities
The following sentences describe the processes of INITIATION, ELONGATION, and TERMINATION. Which one corresponds to which?
In the ________________________________, DNA unwinds and is transcribed into a messenger RNA (mRNA) molecule, which contains a copy of the gene’s base sequence. This mRNA is processed and transported from the nucleus to the cytoplasm, where it will bind to the ribosome to begin the synthesis of the protein.
In the ________________________________, the ribosome reaches a termination codon in the mRNA, indicating that the synthesis of the protein is complete. The polypeptide chain is released from the ribosome and folds into a functional three-dimensional protein.
In the ________________________________, the ribosome moves along the mRNA, reading its base sequence and assembling a chain of amino acids in the correct sequence to form the protein. Amino acids are transported by transfer RNA (tRNA) molecules, which bind to the corresponding codons on the mRNA. The binding of amino acids forms peptide bonds, which create the polypeptide chain of the protein.
2. Indicate the protein formed from the following nucleotide sequence of an RNA:
This gene contains 13 codons of three nucleotides each, and it can be assumed to code for a protein of approximately 13 amino acids. The transcription of this gene would result in a complementary mRNA molecule of 40 nucleotides that would bind to the ribosome to begin the synthesis of the protein. The translation of this gene would follow the sequence of codons and amino acids:
Codon
Amino Acid
AUG
Methionine
GCU
Alanine
CAG
Glutamine
UUC
Phenylalanine
AAG
Lysine
UGG
Tryptophan
AAU
Asparagine
UAC
Tyrosine
AGC
Serine
UCC
Serine
AGU
Serine
GAA
Glutamic acid
UAA
Stop
As it can be observed, the sequence of codons in the mRNA dictates the sequence of amino acids in the resulting protein. In this case, the protein contains Methionine, Alanine, Glutamine, Phenylalanine, Lysine, Tryptophan, Asparagine, Tyrosine, Serine, Serine, Serine, Glutamic acid, and a termination codon.
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3. Given the following DNA nucleotide sequence, obtain the RNA sequence and then carry out the synthesis of the protein
TAC CCT GTT ATA ACG ACT
Extend to see the answer key. [expand]
To obtain the complementary DNA sequence to an RNA sequence, it should be considered that the nitrogenous bases adenine (A) in RNA pairs with thymine (T) in DNA, while guanine (G) in RNA pairs with cytosine (C) in DNA. In addition, uracil (U) in RNA is replaced by thymine (T) in DNA.
Therefore, the complementary DNA sequence to the given RNA sequence would be:
TAC CCT GTT ATG ACG ACT
This gene contains 6 codons of three nucleotides each and it can be assumed to code for a protein of around 6 amino acids. The transcription of this gene would result in a complementary mRNA molecule of 18 nucleotides that would bind to the ribosome to start the synthesis of the protein. The translation of this gene would follow the sequence of codons and amino acids:
There are several sources of simulation and videos about protein synthesis. Here are some suggestions:
The Virtual Cell Animation Collection by North Dakota State University: This website provides a variety of interactive animations and simulations related to cell biology, including protein synthesis. The animations can be accessed for free at https://vcell.ndsu.edu/animations/.
HHMI BioInteractive Protein Synthesis Video: This video, produced by the Howard Hughes Medical Institute, provides a detailed explanation of protein synthesis, including transcription and translation. It can be viewed for free at https://www.biointeractive.org/classroom-resources/protein-synthesis.
Protein Synthesis Simulation by PhET: This simulation allows users to explore the process of protein synthesis by assembling a protein from amino acids and following the steps of transcription and translation. It can be accessed for free at https://phet.colorado.edu/en/simulation/protein-synthesis.
Amoeba Sisters Protein Synthesis Video: This video, produced by the Amoeba Sisters, provides a simplified explanation of protein synthesis, including the role of DNA, RNA, and ribosomes. It can be viewed for free at https://www.youtube.com/watch?v=4SQDyb1nCfo.
If we separate the constituents of the heterogeneous system, we would be carrying out a process of phase separation. See below the procedures by which these phases can be isolated:
Separation methods for heterogeneous systems
Sublimation: allows isolating substances that sublime and vaporize from those that do not. For example, it is possible to separate sand from iodine.
Levigation: allows separating lighter solids from heavier ones. For example, separating rock pieces from sand. This procedure is carried out by currents of water that drag the light particles.
Decantation: is the process by which a solid is separated from a liquid or two immiscible liquids (such as water and oil), by different density.
Centrifugation: allows accelerating the decantation process by means of centrifugal force (circular motion). Decantation will be faster as the number of turns in a given period of time increases.
Extraction: consists of separating one of the components of a mixture by dissolving it with the appropriate solvent. For example, to extract the green pigment from a leaf, it is boiled in a water bath with alcohol for a few minutes.
Another example of phase separation using the extraction method is in the preparation of tea or coffee infusion, extracting from the leaves or beans substances that give the characteristic aroma and flavor.
Solubilization: is the process by which a solid is separated from another by dissolving one of the phases in water. For example, the sugar-sand system.
Filtration: using this procedure, it is also possible to separate a solid phase from a liquid one (such as the water-sand system).
Sieving: is the process by which solids such as sand and rocks are separated, taking advantage of their different volume.
Separation methods for homogeneous systems
Homogeneous systems can be fractionated when they are not pure substances. The methods are as follows:
Distillation: in a homogeneous system of water and alcohol, the difference in boiling points can be used to separate them. It consists of evaporating one of the liquids and then condensing it by cooling. If the boiling points are close, fractional distillation is used, different from simple distillation, used if they are distant points.
There are two main types of distillation: simple distillation and fractional distillation.
Simple distillation is used to separate two liquids with different boiling points. The liquid mixture is heated until it boils, and the vapor rises and is collected in a condenser. The condenser cools the vapor, which turns back into a liquid and is collected in a separate container. This method works well when the boiling points of the two liquids are far apart.
Fractional distillation is used when the boiling points of the two liquids are closer together. In this method, the liquid mixture is heated and the vapor is passed through a column that is packed with a material, such as glass beads or metal mesh, which provides a large surface area for the vapor to condense. As the vapor rises up the column, it cools and condenses, and the liquids with lower boiling points will condense first. The fractions with different boiling points are then collected separately.
Both types of distillation are commonly used in chemistry and industry for separation and purification of liquids.
Crystallization: isolates solids that crystallize from the solution in which they are dissolved. Saltwater is an example. When heated, water evaporates and salt crystals remain. It is a reverse process to dissolution, which is frequently used for substance purification in industry, such as obtaining table salt.
Other separation methods
there are other separation methods in addition to the ones mentioned. Here are some additional separation methods:
Chromatography: This is a technique used to separate and identify the components of a mixture. It involves passing the mixture through a stationary phase, such as a column packed with a solid material, and a mobile phase, such as a liquid or gas that moves through the stationary phase. Different components will move through the stationary phase at different rates and can be collected and analyzed separately.
Electrolysis: This is a process that uses an electrical current to drive a chemical reaction. It can be used to separate mixtures of compounds that can be ionized, such as ionic compounds dissolved in water. The current causes the ions to move toward electrodes of opposite charges, where they can be collected separately.
Precipitation: This method involves adding a chemical to a mixture that causes one or more components to form a solid precipitate that can be separated from the remaining liquid.
Magnetization: This method involves using magnets to separate magnetic materials, such as iron or steel, from non-magnetic materials, such as sand or gravel.
Crystallization by evaporation: This is a process where a solution is allowed to evaporate slowly, causing the dissolved solute to form crystals that can be collected and separated from the remaining liquid.
Membrane filtration: This is a method of separating particles or molecules of different sizes using a membrane with specific pore sizes. The smaller particles or molecules will pass through the membrane, while the larger ones will be retained and can be collected separately.
These are just a few examples of other separation methods. The choice of method will depend on the specific properties of the mixture and the components that need to be separated.
Summary of Separation Methods
Separation Method
What it is used for
Centrifugation
Separating solids from liquids by spinning them rapidly around a central axis. Used to separate mixtures of substances with different densities, such as blood cells from plasma.
Levigation
Separating heavier particles from lighter particles by shaking them in water. Used to separate sand from gravel, for example.
Decantation
Separating a liquid from a solid by pouring the liquid off the top. Used to separate oil and water, for example.
Extraction
Separating a specific substance from a mixture using a solvent. Used to extract caffeine from coffee beans or essential oils from plants.
Sieving
Separating small solid particles from larger ones using a mesh screen. Used to separate flour from lumps or to remove pebbles from sand.
Evaporation
Separating a liquid from a solid by heating it to turn the liquid into a gas, leaving the solid behind. Used to separate salt from water, for example.
Chromatography
Separating a mixture of substances based on their different chemical properties. Used to identify and isolate specific compounds in a mixture, such as different pigments in ink.
Distillation
Separating a mixture of liquids based on their different boiling points. Used to separate alcohol from water in distillation of spirits or to purify water.
Filtration
Separating solid particles from a liquid or gas by passing it through a filter. Used to remove impurities from water, for example.
Crystallization
Separating a solid from a liquid by cooling the solution, causing the solid to form crystals. Used to purify substances such as salt or sugar.
Activities:
1. Indicate the method or methods you would use to separate…
…a system made up of vinegar, water, and oil.
…the system of lemon juice-lemon pulp.
…the salt-water-coal dust system.
…the system of rocks-gold nuggets-river water.
2. What is the difference between filtration and decantation, and when would you use one method over the other?
3. How does centrifugation work, and what type of mixtures is it typically used for?
4. What is distillation, and how is it used to separate a mixture of liquids with different boiling points?
5. How does chromatography work, and what types of mixtures is it commonly used to separate?
6. Explain the process of magnetization, and give an example of a mixture that can be separated using this method.
Teaching resources
There are several online sources where you can find simulations or virtual labs for separating methods of mixtures. Here are a few examples:
PhET Simulations – This website provides a variety of interactive simulations for science and math topics, including simulations for separating mixtures. They offer simulations for chromatography, distillation, and more. You can access their simulations for free at https://phet.colorado.edu/en/simulations/category/chemistry.
Virtual Chemistry Lab – This website provides a virtual lab for separation methods that allows you to perform experiments using a variety of separation techniques, including filtration, decantation, and distillation. You can access the virtual lab for free at http://www.virtual-chemistry-lab.com/separation-of-mixtures.html.
ChemCollective – This website offers a range of interactive activities and simulations for chemistry topics, including separating mixtures. They offer simulations for distillation, chromatography, and more. You can access their resources for free at http://chemcollective.org/vlabs.
Labster – This is a virtual lab platform that offers a range of simulations for science topics, including chemistry. They offer virtual labs for separation methods such as chromatography, filtration, and more. You can access their resources through a subscription at https://www.labster.com/.
CK-12 – This website offers a variety of interactive activities and simulations for science and math topics, including separating mixtures. They offer simulations for distillation, filtration, and more. You can access their resources for free at https://www.ck12.org/student/.
