Specific heat capacity of ice = 2.1 J g^-1 K^-1, specific heat capacity of water = 4.2 J g^-1 K^-1, specific latent heat of fusion of ice = 336 J g^-1.

Solution:

According to the question, 100 g of ice at -10^o C is completely converted into water at 100^o C. And we have : Specific heat capacity of ice = 2.1 J g^-1 K^-1,

specific heat capacity of water = 4.2 J g^-1 K^-1 and specific latent heat of fusion of ice = 336 J g^-1

Using the expression of heat energy i.e., Q = m × c × (change in temperature), we can write the expression for –

Heat energy gained by 100 g of ice at – 10⁰ C in order to raise its temperature to 0⁰ C is

= 100 × 2.1 × 10 = 2100 J

Similarly, heat energy gained by 100 g of ice at 0⁰ C to convert into water at 0⁰ C is

= 100 × 336 = 33600 J

In the same manner, heat energy gained when the temperature of 100 g of water at 0⁰ C rises to 100⁰ C is

= 100 × 4.2 × 100 = 42000 J

Therefore, total amount of heat energy gained is given by the sum of all these energies, i.e.,

= 2100 + 33600 + 42000 = 77700 J

Or, Total energy gain = 7.77 × 10⁴ J