To convert 2 kg of water at 20 °C into ice at 0 °C, liquid ammonia is evaporated. Given specific heat of water = 1 cal/g °C, L(fusion of ice) = 80 cal/g and L(vap of ammonia) = 341 cal/g, the mass of ammonia that must evaporate is closest to:

1. Heat to remove from water (mass 2000 g) to cool 20 °C to 0 °C: $Q_1 = 2000 \times 1 \times 20 = 40000\ \text{cal}$. 2. Heat to remove to freeze it at 0 °C: $Q_2 = m L_{fusion} = 2000 \times 80 = 160000\ \text{cal}$. 3. Total heat absorbed by ammonia: $Q = 40000 + 160000 = 200000\ \text{cal}$. 4. Ammonia evaporated: $m = Q / L_{vap} = 200000 / 341 \approx 586\ \text{g}$. 5. Ignoring the freezing step (only $Q_1$) gives about 117 g, so that trap is rejected. _Source: Balbharati (Maharashtra Board) Class 10 Science & Technology, Ch 5 "Heat", p.81_