How much of heat is to be removed from 50 g of water at 50°C to convert it into ice at 0°C? (Latent heat of fusion of ice = 80 cal g^–1 and specific heat of water = 1 cal/g/°C)

Question

Question

This question was previously asked in

AP Police Constable (Prelims) Official Paper (Held On: 22 Jan 2023)

Answer 1

The Correct answer is 6500 calorie.

Key Points

To calculate the total heat removed from water, we need to consider two stages:
- Cooling the water from 50°C to 0°C.
- Freezing the water at 0°C into ice.
The formula to calculate heat required to cool water is: Q = m × c × ΔT, where:
- m is the mass of the water (given as 50 g).
- c is the specific heat of water (1 cal/g/°C).
- ΔT is the temperature change (50°C - 0°C = 50°C).
Substituting the values: Q = 50 × 1 × 50 = 2500 calories.
For freezing water into ice, the formula is: Q = m × L, where:
- L is the latent heat of fusion of ice (80 cal/g).
Substituting the values: Q = 50 × 80 = 4000 calories.
Total heat removed = 2500 + 4000 = 6500 calories.

Additional Information

Latent Heat of Fusion:
- The latent heat of fusion is the amount of heat energy required to change the state of a substance from solid to liquid (or vice versa) without changing its temperature.
- For water, it is 80 cal/g, which means 80 calories of heat energy are required to convert 1 gram of ice into water at 0°C.
Specific Heat:
- The specific heat is the amount of heat energy required to raise the temperature of 1 gram of a substance by 1°C.
- For water, the specific heat is 1 cal/g/°C, which is relatively high compared to most substances.
Importance of Cooling and Freezing:
- Cooling water to 0°C involves removing heat based on the temperature difference and specific heat.
- Freezing water to ice requires additional heat removal, calculated using the latent heat of fusion.