# Teaching the Laws of Thermodynamics

In this article I demonstrate how I approach thermodynamic problem solutions. I present three problems, all of which are solved by using the first and second laws of thermodynamics. As always, I press the importance of starting the solution of a problem with the statement of the applicable basic rule or principles.

The text editor does not sustain the use of certain notation used in technical literature. As a consequence, I have had to improvise in the following way: (a) Many technical terms are typically represented by Greek letters. I will designate these terms by an italicized capital letter. Specifically for this article, the Greek letter “delta” that is used to represent a change will be written D. (b) Other technical variables will be designated by capital letters with lower case letters serving as subscripts. For example, the heat (Q) removed from a cold (c) reservoir will be written Qc.

Problem. One mole (18 g) of water vaporizes at 373 K and an atmospheric pressure of 1.00 x 10^5 N/m^2. What is its change in internal energy when it vaporizes and expands against atmospheric pressure? Assume the water vapor is an ideal gas.

examination. The quantity in the vapor phase is comparatively large compared to that of the substantial and liquid phases, so the work done when the gas expands is not negligible. From the ideal gas law, the quantity of a mole of water vapor is

……………………………………..Ideal Gas Law

…….V = NRT/P = (1 mol)(8.314 J/molK)(373 K)/(1.00 x 10^5 N/m^2) = 3.10 x 10^-2 m^3.

Since the density of water is 1.0 g/cm^3, the quantity of 1 mol of liquid water is 18 cm^3 = 18 x 10-6 m^3. The change in quantity when the water vaporizes is consequently

………………………DV = (3.10 x 10^-2 m^3 – 18 x 10^-6 m) = 3.10 x 10^-2 m^3.

Notice that the quantity of the water in the liquid state is negligible. The work done when the water vapor expands at continued pressure against the air is now

…………………..W = PDV = (1.00 x 10^5 N/m^2)(3.10 x 10^-2 m^3) = 3.10 x 10^3 J.

The water absorbs heat given by

……………………………Q = mLv= (0.018 kg)(2.26 x 10^6 J/kg) = 4.07 x 10^4 J.

Finally, we find the water’s change in internal energy DU with the first law of thermodynamics:

…………………………………..First Law of Thermodynamics

…………………DU = Q – W = 4.07 x 10^4 J – 3.10 x 10^3 J = 3.76 x 10^4 J.

Problem. A Carnot heat pump operates between the outside air at 278 K and the interior of a house at 296 K. (a) How much heat is depleted to the house for every 1.0 J of work done by the pump? (b) If the house is to be supplied with the same amount of heat when the outside temperature is 268 K, how much work must be done by the pump?

examination. (a) With the first and second laws of thermodynamics, we have

…………..First Law of Thermodynamics……………….Second Law of Thermodynamics

………………W + Qc = Qh…………………………………………..Qc/Tc = Qh/Th

First, we solve the second law equation for Qc in terms of Qh:

…………………………………….Qc = QhTc/Th.

Then we put this into the first law equation and solve for Qh in terms of W:

………………………………….Qh = W/(1 – Tc/Th).

For Tc= 278 K, Th= 296 K, and W = 1.0 J,

……………………….Qh = (1.0 J)/(1 – 278 K/296 K) = 16.4 J.

(b) For the same Qh = 16.4 J and Tc= 268 K, the equation for Qh gives

…………………………..Qh = 16.4 J = W/(1 – 268 K/296 K)

so…………………………W = (1 – 268 K/296 K)16.4 J = 1.6 J.

Notice that the 10 K drop in the outside temperature forces the heat pump to do considerably more work (1.6 J compared with 1.0 J) to get the same amount of heat into the house. For this reason, heat pumps are not nearly as effective in cold climates as they are in more moderate ones.

Problem. Ten kilograms of water at a room temperature of 293 K are placed in a Carnot freezer whose temperature is slightly below 273 K. The freezer exhausts heat to the room, and its compressor operates at a strength of 250 W. How much time is required to freeze the water?

examination. The water is cooled to 273 K, frozen, then left at 273 K. The total heat removed from the water when it is finally frozen is

………………………………Q = Qcool + Qfreeze = mcDT + mLf =

……….(10 kg)(4186 J/kgK)(20 K – 0) + (10 kg)(3.35 x 10^5 J/kg) = 4.19 x 10^6 J.

The work needed to remove this heat from the cold reservoir can be determined with the usual examination of Carnot refrigerators. With the two heat reservoirs at 273 K and 293 K, we have with the second law of thermodynamics:

………………………………Second Law of Thermodynamics

………………………………………Qc/Tc = Qh/Th

………………….4.19 x 10^6 J/273 K = Qh/293 K,

so…………………………………..Qh= 4.50 x 10^6 J.

Now with the first law of thermodynamics,

……………………………….First Law of Thermodynamics

……….W = Qh – Qc = 4.50 x 10^6 J – 4.19 x 10^6 J = 3.1 x 10^5 J.

The motor, operating at 250 W = 250 J/s, can get this much work done in a time t such that

…………………………………..(250 J/s)t = 3.1 x 10^5 J,

so………………….t = (3.1 x 10^5 J)/(250 J/s) = 1.24 x 10^3 s.

There are formulas for efficiency and coefficient of performance that get overused in engine and refrigerator applications. In fact, I use them sometimes. However, students get very little out of the mindless application of these formulas. As much as possible, I try to get my students to solve problems by applying directly the first and secondl laws of thermodynamics.