How much water vapor (100 ° C) should be added to 120 g of ice (0 ° C) so that the final system has a final temperature of 35° C.
2. Relevant equations
Qvapor = lvapor * mvapor during condensation, where lvapor =2260 kJ/kg
Q1 = cwater*mvapor*ΔT during cooling, and cwater = 4,19 kJ/kg
Qmelting =lmelting * mice during melting, and lmelting =333 kJ/Kg
Q2 = cwater*mice*ΔT during heating
3. The attempt at a solution
Energy which is emitted by the vapor is given by: Qvapor + Q1 = 2260m + 4,19*m*100
Energy taken up/used by the ice is given by: Qmelting + Q2 =
333*0,120 + 4,19*0,120*35
I then set energy emitted = energy taken up by the ice and solved for m
2260m + 4,19m*100 = 333*0,120 + 4,19*0,120*35
m= 0,0215 kg or 21,5 g.
Do you think this is the correct way to solve this problem?