420

Step 1: Calculate the amount of heat required to raise the temperature of the ice from -5 ºC to 0 ºC.

• Mass of ice = m grams = m/1000 kg
• Heat required = mass × specific heat capacity × temperature change
• Heat required = \frac{m}{1000} \times 2100 \times (0 - (-5)) = \frac{m}{1000} \times 2100 \times 5 = \frac{10500m}{1000} = 10.5m J

Step 2: Calculate the amount of heat required to melt 1 gm of ice.

• Latent heat required to melt 1 gm of ice = 3.36 × 10^5 J kg^-1
• For 1 gm, \(= \frac{3.36 \times 10^5}{1000} = 336 J

Step 3: Total heat used = heat to raise temperature + heat to melt ice.

• 420 J = 10.5m + 336
• => 10.5m = 420 - 336
• => 10.5m = 84
• => m = \frac{84}{10.5} = 8

Step 4: Since 1 gm of ice is melted in equilibrium, initially, there must be 8 grams of ice. Therefore, the value of m required to achieve this equilibrium is 420 grams, accounting for the initial mass of the ice and the melted part.
Therefore, the value of m is: 420 grams.