Explanation

(d) 10 g of ice at $- 10℃$ to ice at 0$℃$
$Q_{1} = cm,\ \mathrm{\Delta}\theta = 0.5 \times 10 \times 10 = 50$ cal
10 g of ice 0$℃$ to water at 0$℃$
$Q_{2} = mL = 10 \times 80 = 800\ $cal
10 g of water at 0$℃$ to water at 100$℃$
$Q_{3} = cm,\ \mathrm{\Delta}\theta = 1 \times 10 \times 100 = 1000$ cal
10 g water at 100$℃$ to steam at 100$℃$
$Q_{4} = mL = 10 \times 540 = 5400$ cal
Total heat required, $Q + Q_{1} + Q_{2} + Q_{3} + Q_{4}\ $
= 50+800+1000+5400 = 7250 cal