Cover Page Reasons to Study Entropy Intermission Thermodynamic Terms Further Explanations Basic Entropy Calculation Volumes of Gases Partition Added Final Thoughts The End

Basic entropy calculation

When do you use mCvln(Tf-Ti) to find the heat? You use this if the temperature changes. If the temperature doesn’t change you make T constant, but the heat could change under certain circumstances like described earlier. Generally you use the ln formula when the temperature changes and you use mL for fusion changes. Basically you cut up each part of a phase diagram into a certain section and solve for the entropy of each of the sections by themselves and then add them together at the end. This cut and paste method is the only way to get the correct answer. Trying to add steps, like taking the temperature difference of an entire process of several steps and dividing the average heat value of the whole process by this temperature change won’t accomplish anything.

Okay, so how is entropy used?

For our first example we have a 10 g ice cube at -10 C that is put into a lake whose temperature is 15 C. Calculate the change in entropy of the cube-lake system as the ice cube comes to thermal equilibrium with the lake. This is a problem out of the Serway Physics book problems section.

We first assume the final temperature remains constant at 15 C.

Then we use the method of cutting into pieces we talked about earlier. We start with the ice cube:

∆Sice=0.010 kg2220Jkg*Kln273263+0.010 kg333kJkg273K

+ 0.01 kg4180Jkg*Kln288273

Notice there are three parts to the above problem. There is the part where the temperature goes from -10 to 0 C, then there is the heat of fusion of the water section. Then the final section is where the temperature goes from 0 to 10 C.

Now what happens to the entropy of the lake? The lake has constant temperature of 288K.

Slake=(0.010 kg)(2220 J/(kg*K))(273-263 )°CT + (0.010 kg)(333 kJ/kg)T + 0.01 kg4180Jkg*K(288-273)°CT

Here the entropy of the lake uses a different tactic. You don’t use ln terms because the temperature is constant. The change in entropy ∆S is the difference between the entropy lost by the lake and that gained by the ice. This difference is small, at 0.74 J/K.