Question

How much heat (in kj) is evolved in converting 1.00 mol of steam at 130.0 ∘c to ice at -55.0 ∘c? the heat capacity of steam is 2.01 j/g⋅∘c and of ice is 2.09 j/g⋅∘c?

Answer 1

The heat will be evolved in following process:

1) cooling steam from $130^{0}C$ to $100^{0}C$ = Q1

2) conversion of steam to water =Q2

3) Cooling of water to $0^{0}C[/tex=Q3]4) conversion of water to ice=Q45) cooling of ice from [tex]0^{0}C$ to $-55^{0}C$=Q5

Let us calculate each energy one by one:

1) Q1:

Mass of water vapour = moles X molar mass = 1mol X 18 = 18 grams

heat capacity of steam is $2.01\frac{J}{g^{0}C }$

$Q_{1}=massXheat capacityXchangeintemperature=18X2.01X(130-100)=-1085.4J$

2) Q2:

for this we need heat of vaporisation of water = -40.7kJ/mol

so the heat evolved in conversion of stem to water is -40.7kJ

3) Q3

the specific heat of water is $4.18\frac{J}{g^{0}C }$

heat evolved in cooling of water will be $Q_{3}=massXspecific.heatXchange.in.temperature=18X4.184X100=7531.2J=7.531kJ$

4)Q4:

The heat of conversion of ice to water (melting) is 6.02kJ/mol

5) Q5

The cooling of ice will require

$Q_{5}=massXspecificheatXchangeintemperature=18X2.09X(-55)=-2069.1J=-2.069kJ$

The total energy released is

$Q_{total}=Q1+Q2+Q3+Q4+Q5=1.085+40.7+7.531+6.02+2.069=59.474kJ$

The heat released will be 59.474 kJ