Question

Compare the number of calories absorbed when 359 g of ice at 0°C is changed to liquid water at 38°C with the number of calories absorbed when 359 g of liquid water is warmed from 0°C to 38°C.

Answer 1

The number of calories absorbed when 359 g of ice at 0°C is changed to liquid water at 38°C:

$$Q_1=\lambda m+c_wm\Delta t.$$

The number of calories absorbed when 359 g of liquid water is warmed from 0°C to 38°C:

$$Q_2=c_wm\Delta t.$$

The difference:

$$\Delta Q=Q_1-Q_2=\lambda m=80·359=28720\text{ cal}.$$