Question

500.0 g of silver at T = 400K is adde to 300.0 g of water ice at 273.2 K. determine the final temperature of the sytem when it reaches equilibrium. MW H2O= 18.02 g/mol Cp,m[H2O(l)] = 75.291 J/mol K Cp,m[H2O(g)] = 32.12 J/mol K deltaHfus (H2O) = 6.008 KJ/mol deltaHvap (H2O) = 40.656 kJ/mol

Answer 1

Here, the heat flow into the colder water to raise the temperature and

the metal silver heat flows out to cool down.

At the end, both the silver and water will have the same temperature.

Therefore, the heat is flowed out from silver is equal to the heat gained by water.

hence, Qlost = Qgain

ms*Cps*deltaTs = mw*Cpw*deltaTw

500 g x 24.9 J/molK x (400 -Tf) = 3000 g x 32.12 J/mol.K x (Tf-273.2)

(molar heat capcity of silver Cps= 24.9 J/mol.K
  
12450 (400-Tf) = 96360(Tf-273.2)

Tf = 287.71 K

Hence, the final equilibrium temperature is 288 K .