A piece of ice (heat capacity $= 2100\,\text{J kg}^{-1}\,^\circ\text{C}^{-1}$ and latent heat $= 3.36 \times 10^{5}\,\text{J kg}^{-1}$) of mass $m$ grams is at $-5^\circ\text{C}$ at atmospheric pressure. It is given $420\,\text{J}$ of heat so that the ice starts melting. Finally when the ice-water mixture is in equilibrium, it is found that $1\,\text{gm}$ of ice has melted. Assuming there is no other heat exchange in the process, the value of $m$ is

Step 1: Identify the given parameters.

Convert the mass of ice and the mass of melted ice into kilograms.

$$\text{Mass of ice} = m \text{ g} = \frac{m}{1000} \text{ kg}$$ $$\text{Mass of melted ice} = 1 \text{ g} = 10^{-3} \text{ kg}$$ $$\text{Specific heat capacity of ice, } c = 2100 \text{ J kg}^{-1} \text{ }^\circ\text{C}^{-1}$$ $$\text{Latent heat of fusion, } L = 3.36 \times 10^{5} \text{ J kg}^{-1}$$ $$\text{Total heat supplied, } Q = 420 \text{ J}$$

Step 2: Calculate the heat required to raise the temperature of the ice.

The heat $Q_1$ required to raise the temperature of $m$ grams of ice from $-5^\circ\text{C}$ to $0^\circ\text{C}$ is given by:

$$Q_1 = \left(\frac{m}{1000}\right) \cdot c \cdot \Delta T$$ $$Q_1 = \left(\frac{m}{1000}\right) \cdot (2100) \cdot (0 - (-5))$$ $$Q_1 = 10.5m \text{ J}$$

Step 3: Calculate the heat required to melt the ice.

The heat $Q_2$ required to melt $1\text{ g}$ of ice at $0^\circ\text{C}$ is given by:

$$Q_2 = (10^{-3} \text{ kg}) \cdot L$$ $$Q_2 = 10^{-3} \cdot (3.36 \times 10^{5})$$ $$Q_2 = 336 \text{ J}$$

Step 4: Apply the principle of conservation of energy.

The total heat supplied is the sum of the heat used to raise the temperature and the heat used for melting.

$$Q_1 + Q_2 = Q$$ $$10.5m + 336 = 420$$ $$10.5m = 84$$ $$m = \frac{84}{10.5}$$ $$m = 8$$

Answer: The value of $m$ is 8.