Question

Calculate the amount of energy necessary to warm 10.0 g of ice from 0.0° C to 137.0° C. Express your answer in kJ.

Do not enter units of measurement, do not enter the answer in scientific notation.

The specific heat is 4.184 J/g∙°C for water and 1.99 J/g∙°C for steam. ΔHvap is 40.79 kJ/mol and ΔHfus is 6.01 kJ/mol.

Answer 1

Ti = 0.0 oC

Tf = 137.0 oC

here

Lf = 6.01KJ/mol =

6010J/mol

Lets convert mass to mol

Molar mass of H2O = 18.016 g/mol

number of mol

n= mass/molar mass

= 10.0/18.016

= 0.5551 mol

Heat required to convert solid to liquid at 0.0 oC

Q1 = n*Lf

= 0.5551 mol *6010 J/mol

= 3335.9236 J

Cl = 4.184 J/g.oC

Heat required to convert liquid from 0.0 oC to 100.0 oC

Q2 = m*Cl*(Tf-Ti)

= 10 g * 4.184 J/g.oC *(100-0) oC

= 4184 J

Lv = 40.79KJ/mol =

40790J/mol

Heat required to convert liquid to gas at 100.0 oC

Q3 = n*Lv

= 0.5551 mol *40790 J/mol

= 22640.9858 J

Cg = 1.99 J/g.oC

Heat required to convert vapour from 100.0 oC to 137.0 oC

Q4 = m*Cg*(Tf-Ti)

= 10 g * 1.99 J/g.oC *(137-100) oC

= 736.3 J

Total heat required = Q1 + Q2 + Q3 + Q4

= 3335.9236 J + 4184 J + 22640.9858 J + 736.3 J

= 30897 J

= 31.0 KJ

Answer: 31.0 KJ