Question

1 g of ice at 0°C is converted into water by supplying a heat of 418.72 J. The quantity of heat that is used to increase the temperature of water from 0°C is (Latent heat of fusion of ice = \( 3.35 \times 10^5 \, {Jkg}^{-1} \)):

• A. 83.72 J
• B. 33.52 J
• C. 335.72 J
• D. 837.24 J
• E. 418.72 J

Hint:

To calculate the heat used to increase the temperature of water, first subtract the heat used for the phase change (melting of ice) from the total heat supplied.

Answer 1

The Correct Option is A

The total heat supplied in this process has two components:
1. The heat required to melt the ice at 0°C.
2. The heat required to increase the temperature of water from 0°C.
The heat required to melt the ice is given by the formula: \[ Q_{{melt}} = mL, \] where: - \( m = 1 \, {g} = 0.001 \, {kg} \) (mass of ice),
- \( L = 3.35 \times 10^5 \, {J/kg} \) (latent heat of fusion of ice).
So, the heat required to melt the ice is: \[ Q_{{melt}} = 0.001 \times 3.35 \times 10^5 = 335.0 \, {J}. \] 
The total heat supplied is 418.72 J, and the heat required to melt the ice is 335 J. 
Therefore, the remaining heat \( Q_{{water}} \) is used to increase the temperature of the water, which is: \[ Q_{{water}} = Q_{{total}} - Q_{{melt}} = 418.72 \, {J} - 335.0 \, {J} = 83.72 \, {J}. \] Thus, the quantity of heat used to increase the temperature of water from 0°C is 83.72 J, which corresponds to option (A).