Question

At 1 atm, how much energy is required to heat 81.0 g of H2O(s) at –16.0 °C to H2O(g) at 115.0 °C?

Answer 1

Ti = -16.0 oC

Tf = 115.0 oC

here

Cs = 2.09 J/goC

Heat required to convert solid from -16.0 oC to 0.0 oC

Q1 = m*Cs*(Tf-Ti)

= 115 g * 2.09 J/goC *(0--16) oC

= 3845.6 J

Lf = 333.0 J/g

Heat required to convert solid to liquid at 0.0 oC

Q2 = m*Lf

= 115.0g *333.0 J/g

= 38295 J

Cl = 4.184 J/goC

Heat required to convert liquid from 0.0 oC to 100.0 oC

Q3 = m*Cl*(Tf-Ti)

= 115 g * 4.184 J/goC *(100-0) oC

= 48116 J

Lv = 2260.0 J/g

Heat required to convert liquid to gas at 100.0 oC

Q4 = m*Lv

= 115.0g *2260.0 J/g

= 259900 J

Cg = 2.03 J/goC

Heat required to convert vapour from 100.0 oC to 115.0 oC

Q5 = m*Cg*(Tf-Ti)

= 115 g * 2.03 J/goC *(115-100) oC

= 3501.75 J

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

= 3845.6 J + 38295 J + 48116 J + 259900 J + 3501.75 J

= 353658 J

= 353.7 KJ

Answer: 353.7 KJ