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

Question

$Δ_text(vap)H = "40.7 J/mol"$

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Answer 1 · 2017-08-28T11:35:02

$7256.81 kJ$

$DeltaH_(vap) = 40.7kJ*mol^(-1)$
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 $178.3 xx 40.7 = 7256.81 kJ~~73000kJ$