Question

At 1 atm, how much energy is required to heat 43.0 g of H2O(s) at –18.0 °C to H2O(g) at 161.0 °C? Helpful constants can be found here.

Answer 1

Ti = -18.0 oC
Tf = 161.0 oC
here
Cs = 2.09 J/goC

Heat required to convert solid from -18.0 oC to 0.0 oC
Q1 = m*Cs*(Tf-Ti)
= 43 g * 2.09 J/goC *(0--18) oC
= 1617.66 J

Lf = 333.0 J/g

Heat required to convert solid to liquid at 0.0 oC
Q2 = m*Lf
= 43.0g *333.0 J/g
= 14319 J

Cl = 4.184 J/goC

Heat required to convert liquid from 0.0 oC to 100.0 oC
Q3 = m*Cl*(Tf-Ti)
= 43 g * 4.184 J/goC *(100-0) oC
= 17991.2 J

Lv = 2260.0 J/g

Heat required to convert liquid to gas at 100.0 oC
Q4 = m*Lv
= 43.0g *2260.0 J/g
= 97180 J

Cg = 2.01 J/goC

Heat required to convert vapour from 100.0 oC to 161.0 oC
Q5 = m*Cg*(Tf-Ti)
= 43 g * 2.01 J/goC *(161-100) oC
= 5272.23 J

Total heat required = Q1 + Q2 + Q3 + Q4 + Q5
= 1617.66 J + 14319 J + 17991.2 J + 97180 J + 5272.23 J
= 136380 J
= 136 KJ
Answer: 136 KJ