Data: m = 500 g = 0.5 kg,
T1 = 20 °C,
T2 = 100 °C, c = 0.033 kcal/(kg . °C), Q = ?
Q = mc (T2 – T1)
= 0.5 kg × 0.033 kcal/(kg . °C) × (100 °C – 20 °C)
= 0.5 × 0.033 × 80 kcal
= 0.033 × 40 kcal
∴ Q = 1.32 kcal
Heat required = 1.32 kcal.
[Note: 1kcal/(kg-°C) = 1 cal/(g.°C)]