A copper block of mass $$5.0$$ kg is heated to a temperature of $$500°$$C and is placed on a large ice block. What is the maximum amount of ice that can melt?
[Specific heat of copper : $$0.39$$ J g$$^{-1}$$ °C$$^{-1}$$ and latent heat of fusion of water : $$335$$ J g$$^{-1}$$]

The heat released by the copper block as it cools from $$500°$$C to $$0°$$C (temperature of ice) is:

$$Q = mc\Delta T$$

where $$m = 5.0$$ kg $$= 5000$$ g, $$c = 0.39$$ J g$$^{-1}$$ °C$$^{-1}$$, and $$\Delta T = 500°$$C.

$$Q = 5000 \times 0.39 \times 500 = 975000 \text{ J}$$

The mass of ice that can melt using this heat is:

$$m_{ice} = \frac{Q}{L_f}$$

where $$L_f = 335$$ J g$$^{-1}$$.

$$m_{ice} = \frac{975000}{335} = 2910.4 \text{ g} \approx 2.9 \text{ kg}$$

Hence, the correct answer is Option C.