# How much energy is required to vaporize 3210 g of water at 100 °C ?

## Δ_text(vap)H = "40.7 J/mol"

Aug 28, 2017

$7256.81 k J$

#### Explanation:

$\Delta {H}_{v a p} = 40.7 k J \cdot m o {l}^{- 1}$
This means that the total heat required to vaporise 1 mol of liquid water at ${100}^{o} C$ is $40.7 k J$

$1 m o l$ of ${H}_{2} O = 18 g$

$\therefore 3210 g$ of ${H}_{2} O = \frac{3210}{18} = 178.3 m o l$

$1 m o l$ of ${H}_{2} O$ requires $40.7 k J$ of heat.
$\therefore 178.3 m o l$ requires $178.3 \times 40.7 = 7256.81 k J \approx 73000 k J$