Respuesta :
The heat will be evolved in following process:
1) cooling steam from [tex]130^{0}C[/tex] to [tex]100^{0}C[/tex] = Q1
2) conversion of steam to water =Q2
3) Cooling of water to [tex]0^{0}C[/tex=Q3]
4) conversion of water to ice=Q4
5) cooling of ice from [tex]0^{0}C[/tex] to [tex]-55^{0}C[/tex]=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 [tex]2.01\frac{J}{g^{0}C }[/tex]
[tex]Q_{1}=massXheat capacityXchangeintemperature=18X2.01X(130-100)=-1085.4J[/tex]
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 [tex]4.18\frac{J}{g^{0}C }[/tex]
heat evolved in cooling of water will be [tex]Q_{3}=massXspecific.heatXchange.in.temperature=18X4.184X100=7531.2J=7.531kJ[/tex]
4)Q4:
The heat of conversion of ice to water (melting) is 6.02kJ/mol
5) Q5
The cooling of ice will require
[tex]Q_{5}=massXspecificheatXchangeintemperature=18X2.09X(-55)=-2069.1J=-2.069kJ[/tex]
The total energy released is
[tex]Q_{total}=Q1+Q2+Q3+Q4+Q5=1.085+40.7+7.531+6.02+2.069=59.474kJ[/tex]
The heat released will be 59.474 kJ
