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This article shows the solution of four classes of typical calorimetry and thermodynamic problems related to the calculation of the final temperature of a system after carrying out a heat transfer.
- The first case consists of calculating the final temperature of a system, given its heat capacity and the amount of heat absorbed.
- The second is similar to the first, except that the system consists of an ideal gas and the heat capacity is not given.
- The third case combines the principles of thermochemistry with the process learned in case 1. This problem has to do with calculating the final temperature of a calorimeter of known total heat capacity, within which the total combustion of a Known amount of an organic compound.
- Finally, the fourth case is an example of the calculation of the final or equilibrium temperature after the transfer of heat between two bodies that are initially at different temperatures.
In all cases, the calculation is based on the formula that defines the amount of heat:
Where Q represents the amount of heat transferred, C is the heat capacity of the system (also called heat capacity) and DT refers to the temperature change or, what is the same, the difference between the final and initial temperatures.
The formulas for heat capacity in terms of mass and specific heat, as well as moles and molar heat capacity, will also be used.
In these equations m represents the mass, C e the specific heat, n the number of moles and C m the molar heat capacity.
By convention, heat is considered to be positive when it enters the system (causing an increase in temperature) and negative when it leaves the system (causing a decrease in temperature).
Case 1: Calculation of the final temperature of a body after absorbing a known amount of heat.
statement
Determine the final temperature of a copper block that has a total heat capacity of 230 cal/°C and is initially at 25.00°C if it absorbs 7,850 calories as heat from the surroundings.
Solution
In this case, the data available are the initial temperature, the heat capacity and the amount of heat. Also, since the statement specifies that the copper block absorbs heat, then the sign of heat is known to be positive (+). In summary:
Q = + 7,850 cal
C = 230.0 cal/°C
T i = 25.00°C
T f = ?
Now that we have the data sorted, it’s easy to see that all we have to do is solve the second heat equation to get the final temperature, T f . This is achieved by first dividing both members by the heat capacity and then adding the initial temperature to both members:
Now the data is replaced in the equation, it is calculated and that’s it:
Answer
After absorbing 7,850 calories of heat, the copper block heats from 25.00°C to 59.13°C.
Case 2: Calculation of the final temperature of an ideal gas after losing heat.
statement
Determine the final temperature of a sample of air that is initially at a temperature of 180.0 °C occupying a volume of 500.0 L at a pressure of 0.500 atm if it loses 20.021 Joules of heat while keeping the volume constant. Consider air as a diatomic ideal gas for which the molar heat capacity has a value of 20.79 J/mol.K.
Solution
As before, we begin by extracting the data from the statement. The most important thing in this case is to remember that, by convention, the heat leaving the system is negative, so it is essential to be careful not to forget the sign. In addition, you have to be careful with the units, since in this case the heat is given in Jouls and not in calories.
The temperature must also be transformed to Kelvin in order to use the ideal gas law.
T i = 180.0°C + 273.15 = 453.15 K
C m = 20.79 J/mol.K
V = 500.0L
P = 0.500 atm
Q = – 20.021 J
T f = ?
Two additional details are of great importance in this problem. The first is the fact that air can be considered an ideal gas, which implies that the ideal gas law can be used. From this equation (which is presented below), everything is known except for the number of moles, so it can be used to calculate them.
We start by solving the ideal gas law to find the number of moles of air present in the system:
Now you can take two different paths. You can use the moles and molar heat capacity to determine the heat capacity of the system and then use it to calculate the final temperature, or you can combine both equations into one and then solve for T f .
Here we will do the second. First we substitute C = nC m into the heat equation:
Now divide everything by nC m and add the initial temperature in both members, as we did before:
Answer
The air sample is cooled to a temperature of 309.91 K, which is equivalent to 36.76 °C after losing 20.021 J of heat.
Case 3: Calculation of the final temperature of a calorimeter after an exothermic reaction.
statement
A 0.0500 mol sample of benzoic acid, which has an enthalpy of combustion of -3.227, is burned in a constant-pressure calorimeter having a total heat capacity of 4.020 cal/°C and originally at 25°C. kJ/mol. Determine the final temperature of the system when thermal equilibrium is reached.
Solution
n = 0.0500 mol of benzoic acid
∆H c = – 3.227 kJ/mol
C = 4.020 cal/°C
T i = 25.00 °C
T f = ?
In this case, the heat comes from the combustion of benzoic acid. This is an exothermic process (releasing heat) because the enthalpy is negative. However, since the combustion occurs inside the calorimeter, all the heat released by the reaction is absorbed by the calorimeter. This means that:
Where the minus sign reflects the fact that the reaction releases while the system (the calorimeter) absorbs heat, so both heats must have opposite signs.
Also, the heat released by the reaction of 0.500 mol of acid must be the product of the number of moles times the molar enthalpy of combustion:
Therefore, the heat absorbed by the calorimeter will be:
Now, the same equation is used for the final temperature of the first example:
Answer
The calorimeter temperature increases from 25.00 °C to 34.59 °C after the combustion of the benzoic acid sample.
Case 4: Calculation of the final equilibrium temperature by heat transfer between bodies at different initial temperatures.
statement
A 100-g hot piece of iron is introduced into a container with adiabatic walls (which do not conduct heat) containing 250 g of water initially at 15 °C, which is initially at 95 °C. The specific heat of iron is 0.113cal/g.°C.
Solution
In this case there are two systems that are undergoing heat transfer: the water that is in the container and the piece of iron. It should be remembered that the specific heat of water is 1 cal/g.°C. For this reason, the data should be separated by system:
| water data | iron data |
| C e, water = 1 cal/g.°C | C e, iron = 1 cal/g.°C |
| m water = 250 g | m iron = 100 g |
| T i, water = 15.00°C | T i, iron = 95.00°C |
| Tf , water = ? | Tf , iron = ? |
For both water and iron, heat equations can be written:
Where the heat capacity of each system was replaced by the product between its mass and its specific heat. These equations have too many unknowns since we do not know either of the two heats, nor either of the two final temperatures.
Since we have two equations and four unknowns, we need two additional independent equations in order to solve the problem. These two equations consist of the relationship between the two heats and between the two final temperatures.
Since heat flows from one of the systems to the other and we assume that nothing is lost to the surroundings (because the walls are adiabatic) then all the heat released by the iron block is absorbed by the water. Therefore:
Where, again, the negative sign is placed to highlight the fact that one releases heat while the other absorbs it. This sign does not indicate that the heat of water is negative (in fact, it must be positive, since water is the one that absorbs heat), but rather indicates that the sign of the heat of iron is the opposite of that of water. Since the heat of water is positive, then the above equation ensures that the heat of iron is negative, as it is supposed to be.
The other equation relates the final temperatures. Whenever two bodies are in thermal contact, the one with the higher temperature will transfer heat to the cooler one until thermal equilibrium is reached. This occurs when both temperatures are exactly the same. Therefore, the final temperature of both systems must be the same:
Substituting the first two equations in the second, and substituting both final temperatures for T f , we obtain:
In this equation, the only unknown is T f , so all that is left to do is solve it to find that variable. First of all we solve the distributive in both parentheses, then we group terms from the same side and finally we take out the common factor:
Now we substitute the data and voila!
Answer
The equilibrium temperature of the system formed by 250g of water and 100g of iron is 18.46°C.
Tips and recommendations
An important point to keep in mind when carrying out these calculations is that the result should always make sense. If we put two bodies that are at different temperatures in thermal contact, the logical thing is that the final temperature is between both initial temperatures (in this case somewhere between 15°C and 95°C).
If the result is above the higher temperature or below the lower temperature, there must necessarily be an error in the calculations or in the procedure. The most common mistake is to forget to put the minus sign in the equality of the two values.
Another detail to take into account is that the final temperature will always be closer to the initial temperature of the body with the highest heat capacity. In this case, the heat capacity of water is 250 x 1 = 250 cal/°C, while that of iron is 100 x 0.113 = 11.3 cal/°C. As you can see, that of water is more than 20 times higher than that of iron, so it makes sense that the final temperature would be much closer to 15°C, which is the initial temperature of water, than 95°C. which is iron.
References
- Atkins, P., & dePaula, J. (2014). Atkins’ Physical Chemistry (rev. ed.). Oxford, United Kingdom: Oxford University Press.
- Britannica, T. Editors of Encyclopaedia (2018, December 28). Heat capacity . Encyclopedia Britannica. https://www.britannica.com/science/heat-capacity
- Britannica, T. Editors of Encyclopaedia (2021, May 6). Specific heat . Encyclopedia Britannica. https://www.britannica.com/science/specific-heat
- Cedron J.; Landa V.; Robles J. (2011). 1.3.1.- Specific Heat and Heat Capacity | General Chemistry . Retrieved July 24, 2021, from http://corinto.pucp.edu.pe/quimicageneral/contenido/131-calor-especifico-y-capacidad-calorifica.html
- Chang, R. (2008). Physical Chemistry (3rd ed.). New York City, New York: McGraw Hill.
- Chemistry.is. (nd).Specific heat . Retrieved July 24, 2021, from https://www.quimica.es/enciclopedia/Calor_espec%C3%ADfico.html
- Wunderlich, B. (2001). Thermal Analysis. Encyclopedia of Materials: Science and Technology , 9134–9141. https://doi.org/10.1016/b0-08-043152-6/01648-x
