Calorimetry: explanation + 38 exercises with answers

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:

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.

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

1. 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).

Datos:

\( Q=? \)
\( m=23 g \)
\(c=0,115\frac{cal}{g\cdot ^{\circ}C} \)
\( T_{i}=23^{\circ}C \)
\( T_{f}=45^{\circ}C \)

Let’s replace in equation (I):

\( Q=m\cdot c\cdot (T_{f}-T_{i}) \)
\( Q=23g\cdot 0,115\frac{cal}{g\cdot ^{\circ}C}\cdot \left ( 45^{\circ}C-23^{\circ}C \right ) \)
\( Q=58,19cal \)

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:

1. 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:

\( Q_{total}=?\)
\( T_{fus}=0^{\circ}C\)
\(x T_{vap}=100^{\circ}C\)
\( T_{i}=-23^{\circ}C\)
\( T_{f}=130^{\circ}C\)
\( L_{v} =539,4cal/g\)
\( L_{f} =79,7cal/g\)
\( c=1\frac{cal}{g^{\circ}C}\)

Once the data is written, we must analyze the problem.

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:

\( Q_{s}^{1}=m\cdot c\cdot (T_{f}-T_{i})=\)
\( =25g\cdot 1\frac{cal}{g^{\circ}C}\cdot (0^{\circ}C-(-23^{\circ}C))= \)
\( =575 cal\)

\( Q_{L}^{1}=m\cdot L_{f}=25g\cdot 79,7\frac{cal}{g}=1992,5cal\)
\(Q_{s}^{2}=m\cdot c\cdot (T_{f}-T_{i})=\)
\(=25g\cdot 1\frac{cal}{g^{\circ}C}\cdot (100^{\circ}C-0^{\circ}C)=2500 cal\)

\(Q_{L}^{2}=m\cdot L_{v}=25g\cdot 539,4\frac{cal}{g}=13485cal \)
\( Q_{s}^{3}=m\cdot c\cdot (T_{f}-T_{i})=25g\cdot 1\frac{cal}{g^{\circ}C}\cdot (130^{\circ}C-100^{\circ}C)=750 cal \)

Finally, let’s add up all the values of Q obtained:

\( Q_{T}=Q_{s}^{1}+Q_{L}^{1}+Q_{s}^{2}+Q_{L}^{2}+Q_{s}^{3}\)
\(Q_{T}=575cal+1992,5cal+2500cal+13485cal+750cal \)
\( Q_{T}=19302,5cal\)

And so we have finished the example.

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.

\( Q_{2}= -Q_{1} \) (Equation 3)

\(c_{2}\cdot m_{2}.(T_{f}-T_{i_{2}})=-c_{1}\cdot m_{1}.(T_{f}-T_{i_{1}})\)

Now, we only need to solve for Tf to find the final temperature at which thermal equilibrium is reached:

\( T_{f}=\frac{m_{1}\cdot c_{1}\cdot T_{i_{1}}+m_{2}\cdot c_{2}\cdot T_{i_{2}}}{m_{1}\cdot c_{1}+m_{2}\cdot c_{2}}\) (Equation 4)

Example of calorimetry calculations.

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:

\( T_{f}=\frac{m_{1}\cdot c_{1}\cdot T_{i_{1}}+m_{2}\cdot c_{2}\cdot T_{i_{2}}}{m_{1}\cdot c_{1}+m_{2}\cdot c_{2}} \)

\( T_{f}=\frac{10000g\cdot 1\frac{cal}{g\cdot ^{\circ}C}\cdot 13^{\circ}C+90000g\cdot 1\frac{cal}{g\cdot ^{\circ}C}\cdot 27^{\circ}C}{10000g\cdot 1\frac{cal}{g\cdot ^{\circ}C}+90000g\cdot 1\frac{cal}{g\cdot ^{\circ}C}}\)

We solve the calculation and obtain:

\( T_{f}=25.6^{\circ}C \)

Solved Exercises (Yes, With Answers!)

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

1. 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.
2. Express the previous answer in Joules. Ans: 4.18 J.
3. How many J correspond to 234 cal? Ans: 978.12 J.
4. How many cal correspond to 45.6 J? Ans: 10.91 cal.
5. Is cal the same as Cal? Explain.
6. Mark T or F: “34000 cal is equal to 34 Cal, so it is also equivalent to 34 Kcal”
7. 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
8. 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
9. 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
10. 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
11. Express the previous answer in kilojoules. Ans: -2.86 kJ
12. Why is the previous value negative? What does it mean: a) Q<0, b) Q>0, c) Q=0?
13. 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
14. 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
15. 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.
16. 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
17. 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
18. 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
19. 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
20. 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
21. 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.
22. 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
23. 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

Exercises in Calorimetry with Latent Heat

24. 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.
25. What do we call regressive and progressive changes in state of aggregation?
26. Correctly name all the changes of state (regressive and progressive) that occur between a solid, a liquid, and a gas.
27. 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
28. 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
29. 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
30. 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
31. 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.

32. 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
33. 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
34. 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
35. 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
36. 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
37. 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
38. 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.

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Source

Sears and Zemansky. Hugh D. Young, Roger A. Freedman, A. Lewis Ford; “University Physics with Modern Physics 1”; Ed. Pearson Education; available at: https://www.pearsonenespanol.com/mexico/educacion-superior/sears_index/sears-fisica-universitaria-1

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