Explanation

(c) Heat required to convert 10 g of ice at 0$℃$ to water at 0$℃$
$Q_{1}mL = 10 \times 80$ cal
Heat required to raise the temperature of water from 0$℃$ to 20$℃$
$Q_{1} = cm\theta = 1 \times 10 \times 20 = 200$ cal
Total heat required
= $Q_{1} + Q_{2} = 800 + 200 = 1000$ cal