How much energy is required to heat 36.0 g H2O from a liquid at 65°C to a gas at 115°C? The following physical data may be useful.ΔHvap = 40.7 kJ/molCliq = 4.18 J/g°CCgas = 2.01 J/g°CCsol = 2.09 J/g°CTmelting = 0°CTboiling = 100°C

energy required=qnet=87.75kJ

Explanation:

we will do it in three seperate step and then add up those value.

first step is to heat the sample of water upto 100C i. e upto boiling pont. because just after this sample of water started vaporization.

q 1 = m c (T2-T1)

q1 = 36.0 g (4.18 J/gC) (100 - 65 C)

q1 = 5267 J = 5.267kJ

next is to vaporize the sample at 100C

q2 = 36.0 g / 18.0 g/mol X 40.7 kJ/mol

q2 = 81.4 kJ

Finally, heat the steam upto 115C

q3 = m c (T2-T1)

q 3 = 36.0 g (2.01 J/gC) (115-100C)

q3 = 1085 J = 1.085kJ

qnet=q1 + q2 + q3

energy required=qnet=87.75kJ