What is the entropy change when dropping hot metal into cold water?

I think the answer in the book is wrong. Problem posed: "In a specific heat experiment 100 gm of lead (cp = 0.0345 cal/gm C) at 100 C is mixed with 200 gm of water a 20 C. Find the difference in entropy of the system at the end from its value before mixing."

I assumed this was an irreversible process, and the key was the final temperature which I calculated at 294.4 K. Following convention, an intermediate reservoir could be put at the final temperature, and the lead would lose so much heat and entropy, the water would gain a little heat and energy, and in the end, the total entropy would increase by 0.20 cal/deg K.

I have calculated final temp a couple ways, with the same answer, but cannot figure out why the answer is what it is. What am I doing wrong, or is the book answer truly incorrect, a rare occurrence?

• I suspect your mistake is not taking into account the fact that temperature is changing as heat is transferred. By the way, eventually someone is going to come by and close this question unless you modify it to be less specific to your particular homework problem. – octonion Nov 12 '17 at 21:27
• Could you show more of your work? Which equations did you use? – pentane Nov 12 '17 at 21:31
• I used the entropy equation $\Delta S = mC \int \ln (T_f/T_i)$, which applies to reversible processes, but I used an intermediary reservoir at temperature $T_f$ to reduce the temperature reversibly. I brought the lead down to $T_f$ and the water up to $T_f$. The answer is supposed to be $\Delta S = +.20 cal/\deg K$. I get instead $\Delta S = +1.7 cal/\deg K$. – Falsoon Nov 13 '17 at 16:23

I confirm your final temperature, although I get 294.5 K instead of 294.4 K (but who's counting). But, to get the entropy change of each object, you can't just use a single intermediate reservoir at the final temperature. You must contact each object with a continuous sequence of reservoirs, running over the range of temperatures from its initial temperature to its final temperature. So, the heat increments will be $CdT$, and the entropy increments will be $\frac{CdT}{T}$, where C is the heat capacity. And you have to integrate the entropy change over these increments. You perform a separate integration for each object. Then you add up the two entropy changes. This is a simple but accurate way of obtaining the entropy change for reversible paths of each object.
• I got $\Delta S_{lead} =−.82$ cal/K and $\Delta S_{water}=+.92$ cal/K. Then I took the difference, $\Delta S_{sys} =.92−(−82)=1.7$ cal/K. However, I think I agree with you that I should add the two entropies to get $\Delta S_{net} = 0.10$ cal/K, which is what you got. However, the book answer is +0.20 cal/K. So indeed, we would agree that the book answer (Halliday & Resnick) is wrong. I would think others would want to know this. – Falsoon Nov 13 '17 at 17:02