The heat energy required to vapourise `5 kg` of water at `373 K` is

Options

A. `2700 K.cal`
B. `1000 K.cal`
C. `27 K.cal`
D. `270 K.cal`

Explanation

To find the heat energy required to vaporize 5 kg of water at 373 K (which is equivalent to 100 °C), we will use the formula for heat energy related to phase change:

### Step-by-Step Solution:

1. **Identify the Latent Heat of Vaporization**:
The latent heat of vaporization of water is approximately \( L_v = 2260 \, \text{kJ/kg} \) (or \( 2.26 \times 10^6 \, \text{J/kg} \)).

2. **Use the Formula for Heat Energy**:
The heat energy \( Q \) required to vaporize a mass \( m \) of water is given by the formula:
\[ Q = m \cdot L_v \]
where:
- \( Q \) is the heat energy (in Joules),
- \( m \) is the mass of the water (in kg),
- \( L_v \) is the latent heat of vaporization (in J/kg).

3. **Substitute the Values**:
Given that \( m = 5 \, \text{kg} \) and \( L_v = 2.26 \times 10^6 \, \text{J/kg} \):
\[ Q = 5 \, \text{kg} \cdot 2.26 \times 10^6 \, \text{J/kg} \]

4. **Calculate the Heat Energy**:
\[ Q = 5 \cdot 2.26 \times 10^6 = 11.3 \times 10^6 \, \text{J} \]
\[ Q = 11,300,000 \, \text{J} \quad \text{or} \quad 11.3 \, \text{MJ} \]

5. **Final Answer**:
The heat energy required to vaporize 5 kg of water at 373 K is approximately \( 11.3 \, \text{MJ} \).