The opposite faces of a cubical block of iron of cross section 4 square cm are kept in contact with steam and melting ice. Calculate the quantity of ice melted at the end of 10 minutes, k for iron = 0.2 CGS units.

Let x be each side of the cube

As area(A) of each face is 4 ${\mathrm{cm}}^2$

A = ${\mathrm{x}}^2 = 4 {\mathrm{cm}}^2 \mathrm{or} \mathrm{x} = 2 \mathrm{cm}$

Further, with usual notation,

       $({\mathrm{T}}_1-{\mathrm{T}}_2) = (100-0) = {100}^0\mathrm{C}$

$({\mathrm{T}}_1 = \mathrm{Temperature} \mathrm{of} \mathrm{steam}, {\mathrm{T}}_2 = \mathrm{temperature} \mathrm{of} \mathrm{melting} \mathrm{ice})$

k = $0.2 \mathrm{CGS} \mathrm{unit}, \mathrm{t} = 10 \min = 10\times 60\mathrm{s} = 600\mathrm{s}$

As       $\mathrm{Q} = \frac{\mathrm{kA}({\mathrm{T}}_1-{\mathrm{T}}_2)\mathrm{t}}{\mathrm{x}}$

       $\mathrm{Q} = \frac{0.2\times 4\times 100\times 600}{2} = 24,000 \mathrm{cal}$

If m gram of ice be melted with this heat and L be the latent heat of fusion of ice. Q = mL.

or m = $\frac{\mathrm{Q}}{\mathrm{L}} = \frac{24,000}{80}(\mathrm{as} \mathrm{L} = 80 \mathrm{cal}/\mathrm{g})$

or m = 300 g