Heat Exchange and Equilibrium Calculation for Melting Ice

To find the initial mass of the ice piece, $m$ (in grams), we analyze the two stages in which the supplied heat is absorbed:

1. Heating the entire mass of ice from $-5^\circ\text{C}$ to $0^\circ\text{C}$:
   The heat required to raise the temperature of $m\text{ grams}$ (which is $m \times 10^{-3}\text{ kg}$) of ice is given by:
   $Q_1 = M_{\text{ice}} \cdot c_{\text{ice}} \cdot \Delta T$
   $Q_1 = (m \times 10^{-3}\text{ kg}) \times (2100\text{ J kg}^{-1}\text{ }^\circ\text{C}^{-1}) \times (0 - (-5))^\circ\text{C}$
   $Q_1 = m \times 10^{-3} \times 2100 \times 5 = 10.5 m\text{ J}$

2. Melting $1\text{ gm}$ of ice at $0^\circ\text{C}$:
   The mass of the ice melted is $m_{\text{melted}} = 1\text{ gm} = 10^{-3}\text{ kg}$. The heat required for melting this mass is:
   $Q_2 = m_{\text{melted}} \cdot L_f$
   $Q_2 = (10^{-3}\text{ kg}) \times (3.36 \times 10^5\text{ J kg}^{-1}) = 336\text{ J}$

3. Total heat supplied to the system:
   Since there is no other heat exchange, the total heat supplied $Q = 420\text{ J}$ is the sum of $Q_1$ and $Q_2$:
   $Q = Q_1 + Q_2$
   $420 = 10.5 m + 336$
   $10.5 m = 420 - 336$
   $10.5 m = 84$
   $m = \frac{84}{10.5} = 8$

Thus, the value of $m$ is 8.