How much heat is required to change `10g` ice at `0^(@)C` to steam at `100^(@)C`? Latent heat of fusion and vaporisation for `H_(2)O` are `80 cl g^(-1)` and `540 cal g^(-1)`, respectively. Specific heat of water is `1cal g^(-1)`.

Options

A. 7290 cal
B. 5400 cal
C. 7200 cal
D. 8100 cal

Explanation

To find the total heat required to change 10 g of ice at 0°C to steam at 100°C, we need to consider three distinct processes involved in this phase change:

1. **Melting of Ice (0°C to Water at 0°C)**:
- The heat required for this process is given by the formula:
\[ Q_1 = m \times L_f \]
where:
- \( m = 10 \, \text{g} \) (mass of ice)
- \( L_f = 80 \, \text{cal/g} \) (latent heat of fusion for water)
- Thus,
\[ Q_1 = 10 \, \text{g} \times 80 \, \text{cal/g} = 800 \, \text{cal} \]

2. **Heating Water from 0°C to 100°C**:
- The heat required for this process is calculated using:
\[ Q_2 = m \times c \times \Delta T \]
where:
- \( c = 1 \, \text{cal/g°C} \) (specific heat of water)
- \( \Delta T = 100°C - 0°C = 100°C \)
- Thus,
\[ Q_2 = 10 \, \text{g} \times 1 \, \text{cal/g°C} \times 100°C = 1000 \, \text{cal} \]

3. **Vaporization of Water (100°C to Steam at 100°C)**:
- The heat required for this process is given by:
\[ Q_3 = m \times L_v \]
where:
- \( L_v = 540 \, \text{cal/g} \) (latent heat of vaporization for water)
- Thus,
\[ Q_3 = 10 \, \text{g} \times 540 \, \text{cal/g} = 5400 \, \text{cal} \]

Now, we can find the total heat required for all three processes:
\[ Q_{\text{total}} = Q_1 + Q_2 + Q_3 \]
Substituting the values we calculated:
\[ Q_{\text{total}} = 800 \, \text{cal} + 1000 \, \text{cal} + 5400 \, \text{cal} = 7200 \, \text{cal} \]

Thus, the total heat required to change 10 g of ice at 0°C to steam at 100°C is **7200 calories**.