A 5\,g piece of ice at -20^ ^ C is put into m kg of water at 30^ ^...

Heat gained by ice to reach 0^{}^{\circ}\text{C} in the liquid phase consists of the heat required to raise its temperature to 0^{}^{\circ}\text{C} and the latent heat of fusion to melt it.

Using the specific heat of ice s_{ice} = 0.5 \text{ cal/g}^{}^{\circ}\text{C} and latent heat of fusion $L = 80 \text{ cal/g}$, the total heat gained is:

$Q_{gained} = m_{ice} s_{ice} \Delta T + m_{ice} L$

$Q_{gained} = 5 \times 0.5 \times 20 + 5 \times 80$

$Q_{gained} = 50 + 400 = 450 \text{ cal}$

Heat lost by water to cool from 30^{}^{\circ}\text{C} to 0^{}^{\circ}\text{C} is:

$Q_{lost} = m_{water} s_{water} \Delta T$

Since the mass of water is $m \text{ kg} = 1000m \text{ g}$ and s_{water} = 1 \text{ cal/g}^{}^{\circ}\text{C}:

$Q_{lost} = (1000m) \times 1 \times 30 = 30000m \text{ cal}$

By the principle of calorimetry, heat gained equals heat lost:

$30000m = 450$

$m = \dfrac{450}{30000} = 0.015 \text{ kg}$

Answer: $0.015$