Question

Quantity per gram per mole Enthalpy of fusion 333.6 J/g 6010. J/mol Enthalpy of vaporization 2257...

Quantity per gram per mole
Enthalpy of fusion 333.6 J/g 6010. J/mol
Enthalpy of vaporization 2257 J/g 40660 J/mol
Specific heat of solid H2O (ice) 2.087 J/(g·°C) * 37.60 J/(mol·°C) *
Specific heat of liquid H2O (water) 4.184 J/(g·°C) * 75.37 J/(mol·°C) *
Specific heat of gaseous H2O (steam) 2.000 J/(g·°C) * 36.03 J/(mol·°C) *

At 1 atm, how much energy is required to heat 49.0 g of H2O(s) at − 18.0 °C to H2O(g) at 135.0 °C? Heat transfer constants can be found in this table.

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Homework Answers

Answer #1

Ti = -18.0 oC

Tf = 135.0 oC

here

Cs = 2.087 J/goC

Heat required to convert solid from -18.0 oC to 0.0 oC

Q1 = m*Cs*(Tf-Ti)

= 49 g * 2.087 J/goC *(0--18) oC

= 1840.734 J

Lf = 333.6 J/g

Heat required to convert solid to liquid at 0.0 oC

Q2 = m*Lf

= 49.0g *333.6 J/g

= 16346.4 J

Cl = 4.184 J/goC

Heat required to convert liquid from 0.0 oC to 100.0 oC

Q3 = m*Cl*(Tf-Ti)

= 49 g * 4.184 J/goC *(100-0) oC

= 20501.6 J

Lv = 2257.0 J/g

Heat required to convert liquid to gas at 100.0 oC

Q4 = m*Lv

= 49.0g *2257.0 J/g

= 110593 J

Cg = 2.0 J/goC

Heat required to convert vapour from 100.0 oC to 135.0 oC

Q5 = m*Cg*(Tf-Ti)

= 49 g * 2 J/goC *(135-100) oC

= 3430 J

Total heat required = Q1 + Q2 + Q3 + Q4 + Q5

= 1840.734 J + 16346.4 J + 20501.6 J + 110593 J + 3430 J

= 152712 J

= 152.7 KJ

Answer: 152.7 KJ

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