Posted by **melisa** on Saturday, January 7, 2012 at 10:55am.

Find the net change in entropy when 213 g of water at 0.0°C is added to 213 g of water at 96.9°C.

- physics -
**drwls**, Saturday, January 7, 2012 at 11:14am
You have to use absolute temperatures to compute entropy changes.

The mixture will come to a final temperature of

Tf = 96.9/2 = 48.45 C = 321.6 K

because the hot and cold masses are equal.

Initial cold water temperature

Tcold = 273.2 K

Initial hot water temperature

Thot = 370.1 K

Entropy change = cold water entropy gain - hot water entropy loss

The hot water S loss, assuming constant specific heat C, is

Mhot*C*ln(370.1/321.6)= 29.9 cal/degree

The cold water S gain is

Mcold*C*ln(321.6/273.2) = 34.7 cal/deg

The net entropy gain is 4.8 cal/deg

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