Posted by melisa on .
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.
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