80 g of steam at 97 °C is released onto an ice slab at 0 °C. Using L(vap) = 540 cal/g, specific heat of water = 1 cal/g °C and L(fusion) = 80 cal/g, how much ice melts?

1. Steam condenses to water at 97 °C: $Q_a = m L_{vap} = 80 \times 540 = 43200\ \text{cal}$. 2. That water cools 97 °C to 0 °C: $Q_b = m c \Delta T = 80 \times 1 \times 97 = 7760\ \text{cal}$. 3. Total heat to ice: $Q = 43200 + 7760 = 50960\ \text{cal}$. 4. Ice melted: $m_{ice} = Q / L_{fusion} = 50960 / 80 = 637\ \text{g}$. 5. Forgetting the cooling step (only $Q_a$) gives 540 g, so that distractor is rejected. _Source: Balbharati (Maharashtra Board) Class 10 Science & Technology, Ch 5 "Heat", p.79_