Calculate the amount of heat (in calories) required to convert `5 gm` of ice at `0^(@)C` to steam at `100^(@)C`

Options

A. `3100`
B. `3200`
C. `3600`
D. `4200`

Explanation

To calculate the amount of heat required to convert 5 grams of ice at \(0^\circ C\) to steam at \(100^\circ C\), we will break the process down into three main steps:

1. **Melting the ice at \(0^\circ C\) to water at \(0^\circ C\)**:
- The heat required for this step can be calculated using the formula:
\[ Q_1 = m \cdot L_f \]
where:
- \(m = 5 \, \text{g}\) (mass of ice),
- \(L_f = 80 \, \text{kcal/g}\) (latent heat of fusion for ice).
- Therefore:
\[ Q_1 = 5 \, \text{g} \cdot 80 \, \text{kcal/g} = 400 \, \text{kcal} \]

2. **Heating the water from \(0^\circ C\) to \(100^\circ C\)**:
- The heat required for this step can be calculated using the formula:
\[ Q_2 = m \cdot c \cdot \Delta T \]
where:
- \(c = 1 \, \text{kcal/(g} \cdot \text{°C)}\) (specific heat capacity of water),
- \(\Delta T = 100^\circ C - 0^\circ C = 100^\circ C\).
- Therefore:
\[ Q_2 = 5 \, \text{g} \cdot 1 \, \text{kcal/(g} \cdot \text{°C)} \cdot 100^\circ C = 500 \, \text{kcal} \]

3. **Converting water at \(100^\circ C\) to steam at \(100^\circ C\)**:
- The heat required for this step can be calculated using the formula:
\[ Q_3 = m \cdot L_v \]
where:
- \(L_v = 540 \, \text{kcal/g}\) (latent heat of vaporization for water).
- Therefore:
\[ Q_3 = 5 \, \text{g} \cdot 540 \, \text{kcal/g} = 2700 \, \text{kcal} \]

Now, we can find the total heat \(Q\) required for the entire process by summing \(Q_1\), \(Q_2\), and \(Q_3\):
\[ Q = Q_1 + Q_2 + Q_3 = 400 \, \text{kcal} + 500 \, \text{kcal} + 2700 \, \text{kcal} = 3600 \, \text{kcal} \]

Thus, the total amount of heat required to convert 5 grams of ice at \(0^\circ C\) to steam at \(100^\circ C\) is **3600 kcal**.