Question

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

`Δ_text(vap)H = "40.7 J/mol"`

Answer 1

`7256.81 kJ`

Explanation:

`DeltaH_(vap) = 40.7kJ*mol^(-1)`
This means that the total heat required to vaporise 1 mol of liquid water at `100^oC` is `40.7kJ`

`1 mol` of `H_2O = 18 g`

`:. 3210 g` of `H_2O = 3210/18 = 178.3 mol`

`1mol` of `H_2O` requires `40.7kJ` of heat.
`:. 178.3 mol` requires `178.3 xx 40.7 = 7256.81 kJ~~73000kJ`