Find what heat is required to change 20 g of 24 ∘C water to 100∘C steam. Express your answer in calories?

For the change of 24°C of water to 100°C of water :
Mass, m = 20 g
Heat capacity, c = 1 cal/g°C
Temperature raise, ΔT = (100 - 24)°C
Hence, heat required = m c ΔT = 20 × 1 × (100 - 24) cal = 1520 cal

For the change of 100°C water to 100°C ice :
Mass, m = 20 g
Latent heat of evaporation, Lv = 540 cal/g
Hence, heat required = m Lv = 20 × 540 cal = 10800 cal

Total heat required = (1520 + 10800) cal = 12320 cal