1 g of water at `100^(@)C` on heating forms 1g of steam at the same temperature by absorbing ______ cal of energy.

Explanation

To solve the question, we need to determine how much energy is absorbed when 1 gram of water at 100°C is converted into steam at the same temperature.

### Step-by-Step Solution:

1. **Identify the Process**: The question involves the conversion of water to steam, which is a phase change. This process requires energy in the form of latent heat.

2. **Understand Latent Heat**: The latent heat of vaporization is the amount of heat required to convert a unit mass of a substance from liquid to gas at constant temperature. For water, this value is approximately 540 calories per gram.

3. **Use the Formula**: The formula to calculate the energy absorbed during the phase change is:
\[ \text{Energy} = m \times L \]
where \( m \) is the mass of the water (in grams) and \( L \) is the latent heat of vaporization (in calories per gram).

4. **Substitute the Values**:
- Mass \( m = 1 \) gram
- Latent heat of vaporization \( L = 540 \) calories/gram

Now substituting these values into the formula:
\[ \text{Energy} = 1 \, \text{g} \times 540 \, \text{cal/g} = 540 \, \text{cal} \]

5. **Final Answer**: Therefore, the energy absorbed when 1 gram of water at 100°C is converted to steam at the same temperature is **540 calories**.