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, Q_lost = Q_gain

    m_sC_psΔT_s = m_wC_pwΔT_w

   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 C_ps= 24.9 J/mol.K
  
   12450 (400-Tf) = 96360(Tf-273.2)

   Tf = 287.71 K

Hence, the final equilibrium temperature is 288 K .