A copper block of mass 2.5 kg is heated in a furnace to a temperature of ${500}^0\mathrm{C}$ and then placed on a large ice block. What is the maximum amount (approx.) of ice that can melt? (Specific heat of copper = 0.39 $\mathrm{J}/{\mathrm{g}}^0\mathrm{C} \mathrm{latent} \mathrm{heat} \mathrm{of} \mathrm{fusion} \mathrm{of} ice = 335 \mathrm{J}/\mathrm{g})$

Here, mass of copper block,

    M = 2.5 kg = 2.5$\times {10}^3 \mathrm{g}$

Rise in temperature of the copper block,

   $∆\mathrm{T} = {500}^0\mathrm{C}-0^0\mathrm{C} = {500}^0\mathrm{C}$

Specific heat of copper. c = 0.39 $\mathrm{J}/{\mathrm{g}}^0\mathrm{C}$

Latent heat of fusion of water, L = 335 J/g

If m is  the mass of ice melted, applying the principle of calorimetry,

Heat gained by ice = heat lost by copper block

i.e., mL = Mc $\text{∆T}$

or   m = $\frac{\mathrm{Mc} ∆\mathrm{T}}{\mathrm{L}}$

           = $\frac{2.5\times {10}^3\left(\mathrm{g}\right)\times 0.39(\mathrm{J}/{\mathrm{gC}}^0)\times 500\left({\mathrm{C}}^{\mathrm{o}}\right)}{335(\mathrm{J}/\mathrm{g})}$

            = 1455 g = 1.455 = 1.5 kg