Question

The following information is given for ethanol at 1 atm:

Boiling Temp = 78.40°C

Delta H vaporization(78.40°C) = 837.0 J/g

Melting Temp = –114.50°C

Delta H fusion(–114.50°C) = 109.0 J/g
Specific heat gas = 1.430 J/g °C
Specific heat liquid = 2.460 J/g °C

A 45.30 g sample of liquid ethanol is initially at 19.90°C. If the sample is heated at constant pressure (P = 1 atm), ________ kJ of energy are needed to raise the temperature of the sample to 93.20°C.

Answer 1

Ti = 19.9

Tf = 93.2

here

Cl = 2.469 J/g.oC

Heat required to convert liquid from 19.9 oC to 78.4 oC

Q1 = m*Cl*(Tf-Ti)

= 45.3 g * 2.469 J/g.oC *(78.4-19.9) oC

= 6543 J

Lv = 837.0 J/g

Heat required to convert liquid to gas at 78.4 oC

Q2 = m*Lv

= 45.3g *837.0 J/g

= 37916 J

Cg = 1.43 J/g.oC

Heat required to convert vapour from 78.4 oC to 93.2 oC

Q3 = m*Cg*(Tf-Ti)

= 45.3 g * 1.43 J/g.oC *(93.2-78.4) oC

= 959 J

Total heat required = Q1 + Q2 + Q3

= 6543 J + 37916 J + 959 J

= 45418 J

= 45.4 KJ

Answer: 45.4 KJ