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 -10o C is completely converted into water at 100o 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