A piece of metal at 80°C is placed into 1.2 litres of water at 72°C. This thermally
isolated system reaches a final temperature of 76°C. Estimate the overall change
of entropy for this system.
(delta)S = Q/T
The Attempt at a Solution
1) Found Q of water (**C = degrees celsius):
Q = mc(delta)T = 1.2kg (1.00 kcal/kg-C) (4C) = 4.8 kcal
2) Found the Average temp of the water:
T = (76 + 72)/2 = 74C = 347 K
3) Change in Entropy, (delta)S = Q/T = 4.8kcal/347K = .0138 kcal/K
Question: Is the Q (heat) of the metal just -4.8kcal?
I assumed so, and found the change in entropy for the metal to be .0136 kcal/K
Therefore giving me an overall entropy for the entire system of:
(delta)S = .0138 - .0136 = .0002 kcal/K or 0.2 cal/K
This number looks to small to me though. Any help would be appreciated!
Last edited by a moderator: