Q 11-10-040NEETJEE MainMedium
How much heat is needed to convert $10$ g of ice at $-10°$C into steam at $100°$C? (specific heat of ice $= 0.5$ cal/g°C, latent heat of fusion $= 80$ cal/g, latent heat of vaporisation $= 540$ cal/g)
Answer: (C) $7250$ cal
Concept. Add the heat for each stage: warm the ice, melt it, warm the water, boil it.
Step 1. Each stage.
- Ice $-10 \to 0°$C: $10 \times 0.5 \times 10 = 50$ cal.
- Melting: $10 \times 80 = 800$ cal.
- Water $0 \to 100°$C: $10 \times 1 \times 100 = 1000$ cal.
- Vaporisation: $10 \times 540 = 5400$ cal.
Step 2. Total. $50 + 800 + 1000 + 5400 = 7250$ cal.
Answer. $7250$ cal (Option C).
Solution by Sreeraj P, M.Sc Physics