Question

At 1 atm, how much energy is required to heat 39.0 g of H2O(s) at –16.0 °C to H2O(g) at 159.0 °C? Helpful constants can be found here.

Answer 1

Ti = -16.0

Tf = 159.0

here

Cs = 2.09 J/goC

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

Q1 = m*Cs*(Tf-Ti)

= 39 g * 2.09 J/goC *(0--16) oC

= 1304.16 J

Lf = 333.0 J/g

Heat required to convert solid to liquid at 0.0 oC

Q2 = m*Lf

= 39.0g *333.0 J/g

= 12987 J

Cl = 4.184 J/goC

Heat required to convert liquid from 0.0 oC to 100.0 oC

Q3 = m*Cl*(Tf-Ti)

= 39 g * 4.184 J/goC *(100-0) oC

= 16317.6 J

Lv = 2260.0 J/g

Heat required to convert liquid to gas at 100.0 oC

Q4 = m*Lv

= 39.0g *2260.0 J/g

= 88140 J

Cg = 2.03 J/goC

Heat required to convert vapour from 100.0 oC to 159.0 oC

Q5 = m*Cg*(Tf-Ti)

= 39 g * 2.03 J/goC *(159-100) oC

= 4671.03 J

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

= 1304.16 J + 12987 J + 16317.6 J + 88140 J + 4671.03 J

= 123419.79 J

= 123.4 KJ

Answer: 123.4 KJ