Question

In: Chemistry

Calculate the amount of energy (in kJ) necessary to convert 517 g of liquid water from 0°C to water vapor at 192°C.

Calculate the amount of energy (in kJ) necessary to convert 517 g of liquid water from 0°C to water vapor at 192°C. The molar heat of vaporization (H_vap) of water is

40.79 kJ/mol. The specific heat for water is 4.184 J/g ·°C, and for steam is 1.99 J/g ·° C. (Assume that the specific heat values do not change over the range of temperatures in the problem.)

______kJ

Expert Solution

Cl = 4.184 J/g.oC

Heat required to convert liquid from 0.0 oC to 100.0 oC

Q1 = m*Cl*(Tf-Ti)

= 517 g * 4.184 J/g.oC *(100-0) oC

= 216312.8 J

Hv = 40.79KJ/mol =

40790J/mol

Lets convert mass to mol

Molar mass of H2O = 18.016 g/mol

number of mol

n= mass/molar mass

= 517.0/18.016

= 28.6967 mol

Heat required to convert liquid to gas at 100.0 oC

Q2 = n*Hv

= 28.6967 mol *40790 J/mol

= 1170538.9654 J

Cg = 1.99 J/g.oC

Heat required to convert vapour from 100.0 oC to 192.0 oC

Q3 = m*Cg*(Tf-Ti)

= 517 g * 1.99 J/g.oC *(192-100) oC

= 94652.36 J

Total heat required = Q1 + Q2 + Q3

= 216312.8 J + 1170538.9654 J + 94652.36 J

= 1481504 J

= 1482 KJ

Answer: 1.48*10^3 KJ

queen_honey_blossom answered 2 years ago

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