Answer:
Case 1 : Entropy of vaporization of water is [tex]\Delta S=6.048 \times 10^{3}[/tex]
Case 2 :Entropy of fusion of ice is [tex]\Delta S=1.2234 \times 10^{3}[/tex]
Result : Entropy of vaporization of water is greater than entropy of fusion of ice
Explanation:
Entropy of closed system is given by [tex]\Delta S=\frac{\Delta Q}{T}[/tex]
Where [tex]\Delta S [/tex] is change of entropy, [tex]\Delta Q [/tex] is change of heat and T is absolute temperature in kelvin
Case 1 : During vaporization of water
It is said that " 1.00 kg of water at 100 degree C is vaporized and converted to steam at 100 degree C"
Given Heat of vaporization of water is [tex]2256 \times 10^{3}[/tex] J/kg.
[tex]\Delta Q = mL[/tex]
m=1 kg of water
L=[tex]2256 \times 10^{3}[/tex] J/kg.
[tex]\Delta Q = mL = 2256 \times 10^{3}[/tex]
Entropy of closed system is given by [tex]\Delta S=\frac{\Delta Q}{T}[/tex]
Where, T=100+273=373K
[tex]\Delta S=\frac{ 2256 \times 10^{3}}{373}[/tex]
[tex]\Delta S=6.048 \times 10^{3}[/tex]
Thus, Entropy of vaporization of water is [tex]\Delta S=6.048 \times 10^{3}[/tex]
Case 2 : During fusion of ice
It is said that " 1.00 kg of ice at 0 degree C is melted and converted to water at 0 degree C"
Given Heat of fusion of ice is [tex]3.34 \times 10^{5}[/tex] J/kg.
[tex]\Delta Q = mL[/tex]
m=1 kg of ice
L=[tex]3.34 \times 10^{5}=334 \times 10^{3} J/kg.
[tex]\Delta Q = mL = 334 \times 10^{3}[/tex]
Entropy of closed system is given by [tex]\Delta S=\frac{\Delta Q}{T}[/tex]
Where, T=0+273=273K
[tex]\Delta S=\frac{ 334 \times 10^{3}}{373}[/tex]
[tex]\Delta S=1.2234\times 10^{3}[/tex]
Thus, Entropy of fusion of ice is [tex]\Delta S=1.2234 \times 10^{3}[/tex]
Therefore, Entropy of vaporization of water is greater than entropy of fusion of ice
