Question

Work done in converting one gram of ice at -10°C into steam at 100°C is (1 cal = 4.2J) (Specific heat of ice = 0.5 cal/g°C, specific heat of water = 1 cal/g°C, Latent heat of fusion = 80 ca/g; Latent heat of vaporisation = 540 cal/g)

• A. 3045 J
• B. 6056 J
• C. 721 J
• D. 616 J

Answer 1

Answer: (A)

3045 J

Answer 2

The process involves four steps:

1. Heating ice from -10°C to 0°C.
2. Melting ice at 0°C to water at 0°C.
3. Heating water from 0°C to 100°C.
4. Vaporizing water at 100°C to steam at 100°C.

The heat required for each step is calculated as follows:

1. $Q_1 = m \times c_{ice} \times \Delta T_{ice}$ $Q_1 = 1 \text{ g} \times 0.5 \text{ cal/g°C} \times (0°C - (-10°C)) = 5$ cal
2. $Q_2 = m \times L_f$ $Q_2 = 1 \text{ g} \times 80 \text{ cal/g} = 80$ cal
3. $Q_3 = m \times c_{water} \times \Delta T_{water}$ $Q_3 = 1 \text{ g} \times 1 \text{ cal/g°C} \times (100°C - 0°C) = 100$ cal
4. $Q_4 = m \times L_v$ $Q_4 = 1 \text{ g} \times 540 \text{ cal/g} = 540$ cal

The total heat required is the sum of heat required in all steps: $Q_{total} = Q_1 + Q_2 + Q_3 + Q_4$ $Q_{total} = 5 \text{ cal} + 80 \text{ cal} + 100 \text{ cal} + 540 \text{ cal} = 725$ cal

Converting total heat to Joules using 1 cal = 4.2 J: $Q_{total\_Joules} = 725 \text{ cal} \times 4.2 \text{ J/cal} = 3045$ J