THERMAL EQUILIBRIUM: EQUATIONS, APPLICATIONS, EXERCISES - PHYSICAL - 2025

The thermal equilibrium of two bodies that are in thermal contact is the state that is reached after a long enough time for the temperatures of both bodies to equalize.

In thermodynamics, thermal contact of two bodies (or two thermodynamic systems) is understood to be a situation in which the bodies have mechanical contact or are separated but in contact with a surface that allows only the passage of heat from one body to the other (diathermic surface).

Figure 1. After a while the ice and the drink will reach their thermal equilibrium. Source: pixabay

In thermal contact there should be no chemical reaction between the systems in contact. There should only be heat exchange.

Everyday situations in which there is heat exchange occur with systems such as the cold drink and the glass, the hot coffee and the teaspoon, or the body and the thermometer, among many other examples.

When two or more systems are in thermal equilibrium?

The second law of thermodynamics states that heat always goes from the body with the highest temperature to the body with the lowest temperature. Heat transfer ceases as soon as the temperatures equalize and the state of thermal equilibrium is reached.

The practical application of thermal balance is the thermometer. A thermometer is a device that measures its own temperature, but thanks to thermal equilibrium we can know the temperature of other bodies, such as that of a person or animal.

The mercury column thermometer is placed in thermal contact with the body, for example under the tongue, and enough time is waited for the thermal equilibrium between the body and the thermometer to be reached and its reading not to vary further.

When this point is reached, the temperature of the thermometer is the same as that of the body.

The zero law of thermodynamics states that if a body A is in thermal equilibrium with a body C and that same body C is in thermal equilibrium with B, then A and B are in thermal equilibrium even when there is no thermal contact between A and B.

Therefore, we conclude that two or more systems are in thermal equilibrium when they have the same temperature.

Thermal equilibrium equations

We assume a body A with initial temperature Ta in thermal contact with another body B with initial temperature Tb. We also assume that Ta> Tb, then according to the second law the heat is transferred from A to B.

After a while, thermal equilibrium will be reached and both bodies will have the same final temperature Tf. This will have an intermediate value at Ta and Tb, that is, Ta> Tf> Tb.

The amount of heat Qa transferred from A to B will be Qa = Ma Ca (Tf - Ta), where Ma is the mass of body A, Ca the heat capacity per unit mass of A and (Tf - Ta) the temperature difference. If Tf is less than Ta then Qa is negative, indicating that the body A gives up heat.

Similarly for body B we have that Qb = Mb Cb (Tf - Tb); and if Tf is greater than Tb then Qb is positive, indicating that body B receives heat. As body A and body B are in thermal contact with each other, but isolated from the environment, the total amount of heat exchanged must be zero: Qa + Qb = 0

Then Ma Ca (Tf - Ta) + Mb Cb (Tf - Tb) = 0

Equilibrium temperature

Developing this expression and solving for the temperature Tf, the final temperature of thermal equilibrium is obtained.

Figure 2. Final equilibrium temperature. Source: self made

Tf = (Ma Ca Ta + Mb Cb Tb) / (Ma Ca + Mb Cb).

As a particular case, consider the case that bodies A and B are identical in mass and heat capacity, in this case the equilibrium temperature will be:

Tf = (Ta + Tb) / 2 ↔ if Ma = Mb and Ca = Cb.

Thermal contact with phase change

In some situations it happens that when two bodies are placed in thermal contact, the heat exchange causes a change of state or phase in one of them. If this happens, it must be taken into account that during the phase change there is no change in temperature in the body that is modifying its state.

If the phase change of any of the bodies in thermal contact occurs, the concept of latent heat L is applied, which is the energy per unit mass necessary for the change of state:

Q = L ∙ M

For example, to melt 1 kg of ice at 0 ° C, 333.5 kJ / kg is required and this value is the latent heat L of fusion of ice.

During melting, it changes from solid water to liquid water, but that water maintains the same temperature as ice during the melting process.

Applications

Thermal balance is part of daily life. For example, let's examine this situation in detail:

-Exercise 1

A person wants to bathe in warm water at 25 ° C. In a bucket, place 3 liters of cold water at 15 ° C and in the kitchen heat water up to 95 ° C.

How many liters of hot water must he add to the bucket of cold water to have the desired final temperature?

Solution

Suppose A is cold water and B is hot water:

Figure 3. Solution to exercise 3. Source: own elaboration.

We propose the equation of thermal equilibrium, as indicated on the blackboard in figure 3 and from there we solve for the mass of water Mb.

We can obtain the initial mass of cold water because the density of the water is known, which is 1Kg for each liter. That is, we have 3 kg of cold water.

Ma = 3kg

So

Mb = - 3 kg * (25 ° C - 15 ° C) / (25 ° C - 95 ° C) = 0.43 kg

Then 0.43 liters of hot water is enough to finally obtain 3.43 liters of warm water at 25 ° C.

Solved exercises

-Exercise 2

A piece of metal weighing 150 g and with a temperature of 95 ° C is introduced into a container containing half a liter of water at a temperature of 18 ° C. After a while thermal equilibrium is reached and the temperature of the water and the metal is 25 ° C.

Suppose that the container with the water and the piece of metal is a closed thermos that does not allow heat exchange with the environment.

Get the specific heat of the metal.

Solution

First we will calculate the heat absorbed by the water:

Qa = Ma Ca (Tf - Ta)

Qa = 500g 1cal / (g ° C) (25 ° C - 18 ° C) = 3500 calories.

That is the same heat given by the metal:

Qm = 150g Cm (25 ° C - 95 ° C) = -3500 calories.

So we can get the heat capacity of the metal:

Cm = 3500 cal / (150g 70 ° C) = ⅓ cal / (g ° C).

Exercise 3

You have 250 cc of water at 30 ° C. To that water that is in an insulating thermos, 25g of ice cubes are added at 0 ° C, with the purpose of cooling it.

Determine the equilibrium temperature; that is, the temperature that will remain once all the ice has melted and the ice water has heated up to match that of the water in the glass initially.

Solution 3

This exercise can be solved in three stages:

1. The first is the melting of ice that absorbs heat from the initial water to melt and become water.
2. Then the temperature drop in the initial water is calculated, due to the fact that it has given heat (Qced <0) to melt the ice.
3. Finally, the molten water (coming from the ice) must be thermally balanced with the water that existed initially.

Figure 4. Solution to exercise 3. Source: own elaboration.

Let's calculate the heat required for melting ice:

Qf = L * Mh = 333.5 kJ / kg * 0.025kg = 8.338 kJ

Then, the heat given by the water to melt the ice is Qced = -Qf

This heat given by the water lowers its temperature to a value T 'that we can calculate as follows:

T '= T0 - Qf / (Ma * Ca) = 22.02 ° C

Where Ca is the heat capacity of water: 4.18 kJ / (kg ° C).

Finally the original mass of water that is now at 22.02 ° C will give up heat to the mass of molten water from the ice that is at 0 ° C.

Finally, the equilibrium temperature Te will be reached after sufficient time:

Te = (Ma * T '+ Mh * 0 ° C) / (Ma + Mh) = (0.25kg * 22.02 ° C + 0.025kg * 0 ° C) / (0.25kg + 0.025kg).

Finally obtaining the equilibrium temperature:

Te = 20.02 ° C.

-Exercise 4

A 0.5 kg chunk of lead comes out of the furnace at a temperature of 150 ° C, which is well below its melting point. This piece is placed in a container with 3 liters of water at room temperature of 20 ° C. Determine the final equilibrium temperature.

Also calculate:

- Amount of heat delivered by lead to water.

- Amount of heat absorbed by water.

Data:

Specific heat of lead: Cp = 0.03 cal / (g ° C); Specific heat of water: Ca = 1 cal / (g ° C).

Solution

First we determine the final equilibrium temperature Te:

Te = (Ma Ca Ta + Mp Cp Tp) / (Ma Ca + Mp Cp)

Te = 20.65 ° C

Then the amount of heat released by lead is:

Qp = Mp Cp (Te - Tp) = -1.94 x 10³ cal.

The amount of heat absorbed by the water will be:

Qa = Ma Ca (Te - Ta) = + 1.94x 10³ cal.

References

1. Atkins, P. 1999. Physical Chemistry. Omega editions.
2. Bauer, W. 2011. Physics for Engineering and Sciences. Volume 1. Mc Graw Hill.
3. Giancoli, D. 2006. Physics: Principles with Applications. 6th.. Ed Prentice Hall.
4. Hewitt, Paul. 2012. Conceptual Physical Science. 5th. Ed. Pearson.
5. Resnick, R. (1999). Physical. Vol. 1. 3rd Ed. In Spanish. Compañía Editorial Continental SA de CV
6. Rex, A. 2011. Fundamentals of Physics. Pearson.
7. Sears, Zemansky. 2016. University Physics with Modern Physics. 14th. Ed. Volume 1.
8. Serway, R., Jewett, J. (2008). Physics for Science and Engineering. Volume 1. 7th. Ed. Cengage Learning.