Question

How many calories are required to change 225g of ice at 0° C to steam at 100° C?

Answer 1

Case 1 :- First, to melt the ice at 0°C to water at 0°C..

The Latent Heat of Fusion of Ice = 80cal/g
So q1 = 225g x 80cal/g = 18,000cal. (18.0kcal).

Case :-2 To heat the water from 0°C to 100°C (boiling point)...
The Specific Heat of liquid water = 1.0cal/g/°C.
we know q = m * S * ΔT -----(1)

m = mass of substance, S = Specific heat Capacity, ΔT = Temperature change

q2 = 225g x 1cal/g/°C x (100-0)°C
= 22,500cal. (22.5kcal).

Case-3 To vaporise the liquid water to steam at 100°C..
Latent Heat of Vaporisation of water = 540cal/g.
q3 = 225g x 540cal/g = 121,500cal. (121.5kcal).

Total heat required = q1 + q2+q3 = 18kcal + 22.5kcal + 121.5kcal.

= 162kcal (162,000 calories).