Question

Two 20.0-g ice cubes at –20.0 °C are placed into 285 g of water at 25.0 °C. Assuming no energy is transferred to or from the surroundings, calculate the final temperature, Tf, of the water after all the ice melts. heat capacity of H2O(s) is 37.7 J/mol*K heat capacity of H2O(l) is 75.3 J/mol*K enthalpy of fusion of H20 is 6.01 kJ/mol

Answer 1

Qice = mice*Cpice*(0-Tice) + mice*Latent fusion of ice

Qice = 2*20 * 2.02 * (0--20) + 2*20*334 = 14976 J

Qhot water = m*C*(Tf-Ti) = 285*4.184*(Tf-25)

Qcold water = mice*Cpwater*(Tf-0) = 2*20*4.184*(Tf-0)

heat balance

-Qhot = Qcold water + Qice

-285*4.184*(Tf-25) = 14976 + 2*20*4.184*(Tf-0)

-285*4.184*(Tf-25) = 14976 + 2*20*4.184*Tf

-285*4.184*Tf + 25*285*4.184 = 14976 + 2*20*4.184*Tf

Tf(2*20*4.184+285*4.184) =  25*285*4.184-14976

Tf = (25*285*4.184-14976 ) / ((2*20*4.184+285*4.184) ) = 10.90 °C