`10 g` ice at `0^@C` is converted into steam at `100^@C`. Find total heat required . `(L_f = 80 cal//g, S_w = 1cal//g-^@C, l_v = 540 cal//g)`

Explanation

To find the total heat required to convert `10 g` of ice at `0°C` into steam at `100°C`, we will break down the process into three main steps:

1. **Melting the Ice (Heat of Fusion)**: The first step is to convert the ice at `0°C` to water at `0°C`. This requires the heat of fusion.

\[ Q_f = m \cdot L_f \]
Where:
- \( m = 10 \, \text{g} \) (mass of ice)
- \( L_f = 80 \, \text{cal/g} \) (latent heat of fusion)

Substituting the values:
\[ Q_f = 10 \, \text{g} \cdot 80 \, \text{cal/g} = 800 \, \text{cal} \]

2. **Heating the Water (Sensible Heat)**: The second step is to heat the water from `0°C` to `100°C`. This requires sensible heat.

\[ Q_s = m \cdot S_w \cdot \Delta T \]
Where:
- \( S_w = 1 \, \text{cal/g°C} \) (specific heat of water)
- \( \Delta T = 100°C - 0°C = 100°C \)

Substituting the values:
\[ Q_s = 10 \, \text{g} \cdot 1 \, \text{cal/g°C} \cdot 100°C = 1000 \, \text{cal} \]

3. **Vaporizing the Water (Heat of Vaporization)**: The third step is to convert the water at `100°C` to steam at `100°C`. This requires the heat of vaporization.

\[ Q_v = m \cdot L_v \]
Where:
- \( L_v = 540 \, \text{cal/g} \) (latent heat of vaporization)

Substituting the values:
\[ Q_v = 10 \, \text{g} \cdot 540 \, \text{cal/g} = 5400 \, \text{cal} \]

4. **Total Heat Required**: Now, we can find the total heat required by summing all the heats calculated above.

\[ Q_{total} = Q_f + Q_s + Q_v \]
Substituting the values:
\[ Q_{total} = 800 \, \text{cal} + 1000 \, \text{cal} + 5400 \, \text{cal} = 7200 \, \text{cal} \]

Thus, the total heat required to convert `10 g` of ice at `0°C` into steam at `100°C` is **7200 calories** or **7.2 kilocalories**.