Question

Two 20.0-g ice cubes at –19.0 °C are placed into 245 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 H20(s) = 37.7J/(mol*K)

Heat cap of H20(l) 75.3 J/(mol*K)

enthalpy of fusion of H20 = 6.01 kJ/mol

Answer 1

let, assume final temparute is T_f

245 gm of water = 245/18 mol=13.6 mol of water

two cube of 20 gm of ice = 2x20gm=40/18mol=2.222 mol of water

Heat capacity of H20(s) = 37.7J/(mol*K)

Heat cap of H20(l) 75.3 J/(mol*K)

enthalpy of fusion of H20 = 6.01 kJ/mol=6010 j/mol

now, total heat loss of the 245 gm water to be, Q_w=13.6 mol *  75.3(J/mol-K) *(25-T_f)( K)=1024.08(25-T_f) J

as no energy is transferred to or from the surroundings, Q_w amount of heat is reason to melt the ice and become water of T_f°C.

as we know the the fusion temperature of ice is 0°C, we can take whole stage of ice melting in 3 step. where in step on the ice temperature goes down to 0°C from -19°C. then in the 2nd step fusion is take place and ice of 0°C become water of 0°C. then in the 3rd step temparature rises from 0°C to T_f°C.

total heat required for the whole process is, Q_i= 2.222 mol*37.7 (J/mol-K)*(0-(-19))(K)+2.222mol*6010J/mol+2.222mol*75.3 (J/mol-K)*(T_f-0)K = 14945.8386+167.3166T_f J

now equating both Q_w and Q_i, we get, 14945.8386+167.3166T_f =1024.08(25-T_f) (167.3166+1024.08)T_f = 25602-14945.8386 1191.3966T_f = 10656.1614 T_f = 8.9442 °C

So, the final temperature would be 8.9442 °C

NOTE THAT, here the unit of all heat capacities are given per unit of K. but we consider the temperature as  °C. That is because here we consider the temperature differance and temperature differance of °C and temperature difference of K is identical.