Question

How much energy is required to heat 36.0g H2O from a liquid at 65 degrees celsius to a gas at 115 degree celsius?

change in Hvap= 40.7 kJ/mol

Specific heat of water(liquid): Cs= 4.18 J/(g degree Celsius)

Specific heat of water(gas): Cs= 2.01 J/(g degree Celsius)

Tboiling= 100 degree Celsius

Answer 1

Ti = 65.0 oC
Tf = 115.0 oC
here
Cl = 4.18 J/g.oC

Heat required to convert liquid from 65.0 oC to 100.0 oC
Q1 = m*Cl*(Tf-Ti)
= 36 g * 4.18 J/g.oC *(100-65) oC
= 5266.8 J

Hvap = 40.7KJ/mol =
40700J/mol


Lets convert mass to mol
Molar mass of H2O = 18.016 g/mol
number of mol
n= mass/molar mass
= 36.0/18.016
= 1.9982 mol

Heat required to convert liquid to gas at 100.0 oC
Q2 = n*Hvap
= 1.9982 mol *40700 J/mol
= 81327.7087 J

Cg = 2.01 J/g.oC

Heat required to convert vapour from 100.0 oC to 115.0 oC
Q3 = m*Cg*(Tf-Ti)
= 36 g * 2.01 J/g.oC *(115-100) oC
= 1085.4 J

Total heat required = Q1 + Q2 + Q3
= 5266.8 J + 81327.7087 J + 1085.4 J
= 87680 J
= 87.7 KJ
Answer: 87.7 KJ