Q.3 (ii) How much heat is required to convert 500 g of ice at 0 °C into water at 40 °C ? (Specific latent heat of fusion of ice = 80 cal/g; specific heat of water = 1 cal/g °C)
Best answer
Answer: 60,000 cal (60 kcal).
Given: m = 500 g, L = 80 cal/g, c = 1 cal/g °C, rise in temperature ΔT = 40 − 0 = 40 °C.
Step 1: Heat to melt the ice at 0 °C (no change in temperature):
\[Q_1 = mL = 500 \times 80 = 40{,}000\ \text{cal}\]
Step 2: Heat to raise the water from 0 °C to 40 °C:
\[Q_2 = mc\,\Delta T = 500 \times 1 \times 40 = 20{,}000\ \text{cal}\]
Total heat: \(Q = Q_1 + Q_2 = 40{,}000 + 20{,}000 = 60{,}000\ \text{cal} = 60\ \text{kcal}\).
Log in or create a free account to answer this question.
