Steam at$\text{100°C}$  required to just melt 3200 g of ice at$-\text{10°C}$ ... (specific heat of ice = 0.5 and latent heat of fusion of ice =$\text{80}\text{cal/g}$ )

t = $\text{100°C}$ , mass of steam = x

m = 3200g of ice; t = – 10 °C

heat liberated as steam at $\text{100°C}$  condenses into water at$\text{100°C}$

${\text{Q}}_{\text{1}}=x\times 540$

Heat absorbed by ice = heat absorbed for melting of ice + heat absorbed for rainsing temperature from $-\text{10°C}\text{to}\text{0°C}$

${\text{Q}}_{\text{2}}=3200\times 0.5\times 10=3200\times 5$

${\text{Q}}_{\text{3}}=3200\times 80$

$\Rightarrow {\text{Q}}_{\text{1}}={\text{Q}}_{\text{2}}+{\text{Q}}_{\text{3}}$

$540x=3200\times 5+3200\times 80$

$x=\frac{16000+2,56,000}{540}=\frac{272000}{540}$

$x=503.7\text{g}$