Question

How much heat (in kJ) is required to warm 13.0 g of ice, initially at -12.0 ∘C, to steam at 111.0 ∘C? The heat capacity of ice is 2.09 J/g⋅∘C and that of steam is 2.01 J/g⋅∘C.

Answer 1

Ti = -12.0

Tf = 111.0

here

Cs = 2.09 J/goC

Heat required to convert solid from -12.0 to 0.0

Q1 = m*Cs*(Tf-Ti)

= 13 g * 2.09 J/goC *(0--12) oC

= 326.04 J

Lf = 333.0 J/g

Heat required to convert solid to liquid at 0.0

Q2 = m*Lf

= 13.0g *333.0 J/g

= 4329 J

Cl = 4.184 J/goC

Heat required to convert liquid from 0.0 to 100.0

Q3 = m*Cl*(Tf-Ti)

= 13 g * 4.184 J/goC *(100-0) oC

= 5439.2 J

Lv = 2260.0 J/g

Heat required to convert liquid to gas at 100.0

Q4 = m*Lv

= 13.0g *2260.0 J/g

= 29380 J

Cg = 2.01 J/goC

Heat required to convert vapour from 100.0 to 111.0

Q5 = m*Cg*(Tf-Ti)

= 13 g * 2.01 J/goC *(111-100) oC

= 287.43 J

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

= 326.04J + 4329J + 5439.2J + 29380J + 287.43 J

= 39762 J

= 39.8 KJ

Answer: 39.8 KJ