The user wants me to solve a physics problem involving heat transfer and phase change.

Here's the plan:

1. Identify the given information:

   • Specific heat capacity of ice, $c_{ice} = 2100 \text{ J kg}^{-1} \text{ }^\circ\text{C}^{-1}$.
   • Latent heat of fusion of ice, $L_f = 3.36 \times 10^5 \text{ J kg}^{-1}$.
   • Mass of the ice piece is $m$ grams.
   • Initial temperature of ice, $T_i = -5 \text{ }^\circ\text{C}$.
   • Total heat supplied, $Q_{total} = 420 \text{ J}$.
   • Mass of ice that melted, $m_{melted} = 1 \text{ g}$.
   • The final state is an ice-water mixture in equilibrium. This implies the final temperature is $0 \text{ }^\circ\text{C}$.

2. Break down the heat transfer process:

   • The total heat supplied ($Q_{total}$) is used for two processes:
     1. Heating the entire piece of ice of mass $m$ from $-5 \text{ }^\circ\text{C}$ to its melting point, $0 \text{ }^\circ\text{C}$. Let's call this heat $Q_1$.
     2. Melting a portion of the ice ($1$ g) at $0 \text{ }^\circ\text{C}$. Let's call this heat $Q_2$.

3. Formulate the equations for each step:

   • Step 1: Heating the ice to 0°C

     • The heat required is given by $Q_1 = m_{total} \cdot c_{ice} \cdot \Delta T$.
     • Here, $m_{total} = m$ grams. I need to be careful with units. The specific heat is given in J kg⁻¹ °C⁻¹. Let's convert the mass $m$ to kg.
     • $m_{total} = \frac{m}{1000}$ kg.
     • The change in temperature is $\Delta T = T_f - T_i = 0 - (-5) = 5 \text{ }^\circ\text{C}$.
     • So, $Q_1 = \left(\frac{m}{1000}\right) \cdot (2100) \cdot (5)$.
     • $Q_1 = m \cdot \frac{2100 \times 5}{1000} = m \cdot (2.1 \times 5) = 10.5 m \text{ J}$.

   • Step 2: Melting 1g of ice at 0°C

     • The heat required for melting is given by $Q_2 = m_{melted} \cdot L_f$.
     • Here, $m_{melted} = 1 \text{ g} = \frac{1}{1000}$ kg.
     • The latent heat is $L_f = 3.36 \times 10^5 \text{ J kg}^{-1}$.
     • So, $Q_2 = \left(\frac{1}{1000}\right) \cdot (3.36 \times 10^5) = \frac{336000}{1000} = 336 \text{ J}$.

4. Set up the total heat equation:

   • The total heat supplied is the sum of the heat used in both steps.
   • $Q_{total} = Q_1 + Q_2$.
   • $420 = 10.5 m + 336$.

5. Solve for m:

   • $420 - 336 = 10.5 m$.
   • $84 = 10.5 m$.
   • $m = \frac{84}{10.5}$.
   • $m = \frac{840}{105}$.
   • Let's simplify this fraction. Both are divisible by 5. $840/5 = 168$. $105/5 = 21$.
   • $m = \frac{168}{21}$.
   • $21 \times 8 = 168$. So, $m = 8$.

6. Final check:

   • The value of $m$ is 8 grams.
   • Let's re-calculate the heat required.
   • Mass of ice = 8 g = 0.008 kg.
   • Heat to raise temperature from -5°C to 0°C: $Q_1 = mc\Delta T = (0.008 \text{ kg}) \times (2100 \text{ J kg}^{-1} \text{ }^\circ\text{C}^{-1}) \times (5 \text{ }^\circ\text{C}) = 0.008 \times 10500 = 8 \times 10.5 = 84 \text{ J}$.
   • Mass of ice melted = 1 g = 0.001 kg.
   • Heat to melt 1g of ice: $Q_2 = m_{melted} L_f = (0.001 \text{ kg}) \times (3.36 \times 10^5 \text{ J kg}^{-1}) = 336 \text{ J}$.
   • Total heat supplied = $Q_1 + Q_2 = 84 \text{ J} + 336 \text{ J} = 420 \text{ J}$.
   • This matches the given total heat. So the calculation is correct.