Question

At 1 atm, how much energy is required to heat 69.0 g of H2O(s) at –16.0 °C to H2O(g) at 141.0 °C?

Answer 1

Ti = -16.0
Tf = 141.0
here
Cs = 2.09 J/goC

Heat required to convert solid from -16.0 oC to 0.0 oC
Q1 = m*Cs*(Tf-Ti)
= 69 g * 2.09 J/goC *(0--16) oC
= 2307.36 J

Lf = 333.0 J/g

Heat required to convert solid to liquid at 0.0 oC
Q2 = m*Lf
= 69.0g *333.0 J/g
= 22977 J

Cl = 4.184 J/goC

Heat required to convert liquid from 0.0 oC to 100.0 oC
Q3 = m*Cl*(Tf-Ti)
= 69 g * 4.184 J/goC *(100-0) oC
= 28869.6 J

Lv = 2260.0 J/g

Heat required to convert liquid to gas at 100.0 oC
Q4 = m*Lv
= 69.0g *2260.0 J/g
= 155940 J

Cg = 2.03 J/goC

Heat required to convert vapour from 100.0 oC to 141.0 oC
Q5 = m*Cg*(Tf-Ti)
= 69 g * 2.03 J/goC *(141-100) oC
= 5742.87 J

Total heat required = Q1 + Q2 + Q3 + Q4 + Q5
= 2307.36 J + 22977 J + 28869.6 J + 155940 J + 5742.87 J
= 215837 J
= 215.8 KJ