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## Homework Statement

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.

## Homework Equations

(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!

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