Question

How much energy is required to change a 40 g ice cube from ice at -35°C to steam at 110°C?

Answer 1

We know that

Specific heat of ice and steam= 0.5 kcal/kg°c

Latent heat of melting = 80 kcal/kg°c

Latent heat of vaporization= 540 kcal/kg°c

To raise the temperature of the ice to 0 ^oC we need

ΔQ = 0.04 kg*(0.5 kcal/(kg^oC))*35 ^oC = 0.7 kcal.
To melt the ice we need
ΔQ = 0.04 kg*80 kcal/kg = 3.2 kcal.
To raise the temperature of the water to 100 ^oC we need
ΔQ = 0.04 kg*(1 kcal/(kg^oC))*100 ^oC= 4 kcal.
To boil the water we need
ΔQ = 0.04 kg*540 kcal/kg = 21.6 kcal.
To raise the temperature of the steam to 110^oC we need
ΔQ = 0.04kg*(0.5kcal/(kg^oC))*10 ^oC = 0.2 kcal.
The total thermal energy required is
(0.7 + 3.2 + 4 + 21.6 + 0.2)kcal = 29.7 kcal.