How much heat is required to change 5 gram ice `(0^@C)` to steam at `100^@C` ? Latent heat of fusion and vaporization for water are 80 cal/g and 540 cal/g respectively . Specific heat of water is 1 cal/ g/k.

Options

A. 7200 cal
B. 3600 cal
C. 1800 cal
D. 900 cal

Answer

To calculate the total heat required to change 5 grams of ice at 0°C to steam at 100°C, we will break the process down into three steps:

1. **Melting the Ice to Water at 0°C (Q1)**:
- The heat required to melt ice (latent heat of fusion) is given by the formula:
\[ Q_1 = m \times L_f \]
where:
- \( m = 5 \, \text{g} \) (mass of ice)
- \( L_f = 80 \, \text{cal/g} \) (latent heat of fusion)

Substituting the values:
\[ Q_1 = 5 \, \text{g} \times 80 \, \text{cal/g} = 400 \, \text{cal} \]

2. **Heating the Water from 0°C to 100°C (Q2)**:
- The heat required to raise the temperature of water is given by the formula:
\[ Q_2 = m \times S \times \Delta T \]
where:
- \( S = 1 \, \text{cal/g/K} \) (specific heat of water)
- \( \Delta T = T_f - T_i = 100°C - 0°C = 100 \, \text{K} \)

Substituting the values:
\[ Q_2 = 5 \, \text{g} \times 1 \, \text{cal/g/K} \times 100 \, \text{K} = 500 \, \text{cal} \]

3. **Vaporizing the Water to Steam at 100°C (Q3)**:
- The heat required to vaporize water (latent heat of vaporization) is given by the formula:
\[ Q_3 = m \times L_v \]
where:
- \( L_v = 540 \, \text{cal/g} \) (latent heat of vaporization)

Substituting the values:
\[ Q_3 = 5 \, \text{g} \times 540 \, \text{cal/g} = 2700 \, \text{cal} \]

4. **Total Heat Required (Q_total)**:
- Now, we can find the total heat required by summing up \( Q_1 \), \( Q_2 \), and \( Q_3 \):
\[ Q_{\text{total}} = Q_1 + Q_2 + Q_3 \]
Substituting the values:
\[ Q_{\text{total}} = 400 \, \text{cal} + 500 \, \text{cal} + 2700 \, \text{cal} = 3600 \, \text{cal} \]

Thus, the total heat required to change 5 grams of ice at 0°C to steam at 100°C is **3600 calories**.