Question

How much heat is required to convert 1g of ice at 0 degree Celsius to steam at 100 degree Celsius (Heat of fusion of ice = 335 x 10^3 Jkg^-1, heat of vaporization of water = 2.26 x 10^6JKg^-1, specific heat capacity of water = 4200Jkg^-1K^-1)

Answer 1

The total heat consists of the heat of ice melting, water heating and vaporization:

$$Q=Q_m+Q_h+Q_v$$

the heat of ice melting:

$$Q_m=\lambda m$$

Heat of water heating:

$$Q_h=Cm(T_2-T_1)$$

And heat of vaporization:

$$Q_v=Lm$$

Then

$$Q=(\lambda+L+C(T_2-T_1))m=\\ =(335\times10^3+2.26\times10^6+4200\times100)\times0.001 = 3015J$$