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
$$Q = 10(0.5)(10) + 10(80) + 10(1)(100) + 10(540) = 50 + 800 + 1000 + 5400 = 7250\ \text{cal}$$
Solution by Sreeraj P, M.Sc Physics