A thermally insulated pot holds 150 g of ice at 0 °C. How much steam at 100 °C must be added so that the final mixture is water at 50 °C? (L(fusion) = 80 cal/g, L(vap) = 540 cal/g, specific heat of water = 1 cal/g °C)

1. Heat gained by ice = melt + warm to 50 °C: $150 \times 80 + 150 \times 1 \times 50 = 12000 + 7500 = 19500\ \text{cal}$. 2. Let steam mass be m. Heat lost = condense + cool 100 °C to 50 °C: $m \times 540 + m \times 1 \times 50 = 590m$. 3. Principle of heat exchange: $590m = 19500$. 4. $m = 19500 / 590 \approx 33\ \text{g}$. 5. Dropping the steam's cooling term (divide by 540 only) gives about 36 g; dropping the ice's warming term gives 19 g, both rejected. _Source: Balbharati (Maharashtra Board) Class 10 Science & Technology, Ch 5 "Heat", p.81_