1.) To convince yourself that entropy of a bath changes even though its temperat
ID: 788833 • Letter: 1
Question
1.) To convince yourself that entropy of a bath changes even though its temperature remains
unchanged, consider the following case: A tub that contains water at 40C receives 100 kJ of
heat. Calculate the final temperature of the water and its entropy change if the mass of the
water in the tub is:
a.) 1 kg
b.) 10 kg
c.) 1000 kg
d.) Compare your results to the calculation ?S = Qbath/Tbath
e.) Prove that when the mass of the bath approaches infinity the entropy change of the bath is
?Sbath = Cp(T2-T1)/T2 Hint: Recall that ln
I got the answers but want to double sure.
Explanation / Answer
To start, be sure that all of your temperatures are in Kelvin. Thermo needs to have absolute temperatures, not relative ones. You can sometimes get away with C, depending on your calculations, but K makes it much easier. I use C for finding the new temperatures, since a difference between two temps in C is the same as in K. However, for finding change in entropy, I use K, since the ratio of two temps in K is different than the same two temps in C.
For a-c, recall that Q = m*Cp*(Tf-Ti). You can rearrange this to be Ti + Q/m/C = Tf.
Since Cp for water is 4.184 KJ/Kg/K, we get
a) 63.901 C
b) 42.390 C
c) 40.024 C
For the change in entropy, we are at constant pressure, so we use deltaS = m*Cp*ln(Tf/Ti) (we use m instead of n since Cp is in units of KJ/Kg/K, instead of KJ/mol/K). We get
a) 0.307 KJ/K
b) 0.318 KJ/K
c) 0.319 KJ/K
d) Here, we use the temperature that the water started at. We get 100 KJ/40 K = 0.319 KJ/K. This is the case for a thermodynamically reversible process (no entropy generated by gradients and such). Note that as the bath gets larger, the change in entropy approaches the change in entropy for the reversible process!
e) Oh snap, that's kinda cool! Not sure how to do that last bit. Its been a while since I took this class. How did you do that?
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