How much heat must be supplied to convert 10 g of ice at 0 °C into water at 50 °C? (Latent heat of fusion = 80 cal/g, specific heat of water = 1 cal/g °C)

1. First melt the ice: $Q_1 = m L_{melt} = 10 \times 80 = 800\ \text{cal}$. 2. Then heat the water from 0 °C to 50 °C: $Q_2 = m c \Delta T = 10 \times 1 \times 50 = 500\ \text{cal}$. 3. Total: $Q = Q_1 + Q_2 = 800 + 500 = 1300\ \text{cal}$. 4. Omitting the melting step gives only 500 cal, so that distractor is rejected. _Source: Balbharati (Maharashtra Board) Class 10 Science & Technology, Ch 5 "Heat", p.72_