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

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