## Melting Ice

0.100 kg of water at 89.0°C is poured into an insulated cup containing 0.460 kg of ice initially at 0°C. How many kg of liquid will there be when the system reaches thermal equilibrium?

**0.212 kg**## **Approach**

The amount of ice (0.460 kg) is significantly larger than the amount of water (0.100 kg) and also the latent heat of fusion for water is significantly larger than the heat required to raise the temperature of water per °C. Thus not all the ice will melt. The final amount of water that is in the cup is the amount of ice that has melted plus the original amount of water.

**Solution**

Let m1 be the amount of ice that has melted. By the principle of energy conservation

miLf=mwater*cwater*dTemp

where the latent heat of fusion for water is Lf = 333 kJ/kg and specific heat of water, cwater = 4.186 kJ/(kg·K), and ΔT = 89 - 0 = 89°C. Putting in the numbers, the amount of ice that has melted is 0.1119 kg. The total amount of water is then 0.1119 + 0.1 kg = 0.212 kg.

miLf=mwater*cwater*dTemp

where the latent heat of fusion for water is Lf = 333 kJ/kg and specific heat of water, cwater = 4.186 kJ/(kg·K), and ΔT = 89 - 0 = 89°C. Putting in the numbers, the amount of ice that has melted is 0.1119 kg. The total amount of water is then 0.1119 + 0.1 kg = 0.212 kg.