Given: Specific heat capacity of water = 4200 J kg^-1 K^-1, specific latent heat of ice = 336 × 10³ J Kg^-1.

Solution:

According to the question, 2 kg of ice melts when water at 100^oC is poured in a hole drilled in a block of ice. We have to find the mass of water used in this process. We have,

Specific heat capacity of water = 4200 J kg^-1 K^-1 and specific latent heat of ice = 336 × 10³ J Kg^-1.

The final temperature would be 0⁰ C because the whole block does not melt and only 2 kg of it melts. We first find the amount of heat energy involved in the process by using the expression below –

Q = m × c × (change in temperature)

Using the above expression, amount of heat energy gained by 2 kg of ice at 0⁰ C in order to convert into water at 0⁰ C is

= 2 × 336000 = 672000 J

Suppose the amount of water poured is m kg

Then, the initial temperature of water is 100⁰ C and we stated the final temperature of water above, i.e., 0⁰ C. Now, using the expression of heat energy amount of heat energy lost by m kg of water at 100⁰ C to reach temperature 0⁰ C =is

= m × 4200 × 100 = 420000 m J

We know that heat energy gained is equal to the heat energy lost. so, we have :

672000 J = m × 420000 J

Or, m = 672000 / 420000

Therefore, m = 1.6 kg