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?

The heat will be evolved in following process:

1) cooling steam from to = Q1

2) conversion of steam to water =Q2

3) Cooling of water to to =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) 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

heat evolved in cooling of water will be

4)Q4:

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

5) Q5

The cooling of ice will require

The total energy released is

The heat released will be 59.474 kJ

when the volume of gas is   changed from   717.73 ml   to 852 ml   the temperature will   change   from 315   c to 452   c

    calculation

this is calculated using the charles law formula

that is,   v1/t1 = v2/t2

          v1=?

        t1 = 315 c   into   kelvin = 315 +273 = 588 k

        v2 = 852   ml

        t2 = 425c + 273 =698 k

make v1   the subject of the formula ( by   multiplying both side by t1)

  v1=( v2 xt1) / t2

v1= (852 ml   x 588 k)/ 698 k= 717.73 ml