Question

The latent heat of vaporization of water is 2.26x106
J/kg. How many grams of water at 100°C can be converted to steam by 2.26x106
J of energy

Answer 1

Answer : 1000 grams of water at 100 °C can be converted to steam by 2.26 × 10⁶ J of energy.

Explanation:

Given : latent heat of vaporization ( Lvap ) = 2.26 × 10⁶ J/kg

Heat required to convert water into steam =. 2.26 × 10⁶ J

To find : grams of water ( m ) = ?

In this case no temperature change takes place only change of state occurs.

Formula : Heat Required = m × Lavp

Substituting the given value in above formula :

∴ 2.26 × 10⁶ = m × 2.26 × 10⁶

∴ m = 2.26 × 10⁶/ 2.26 × 10⁶

∴ m = 1kg

∴ m = 1000 grams.