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

#Δ_text(vap)H = "40.7 J/mol"#

1 Answer
Aug 28, 2017

Answer:

#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#