Calculate the change in entropy when 1.00 kg of water at 100 degree c is vaporized and converted to steam at 100 degree c. assume that the heat of vaporization of water is 2256 times 103 j/kg. calculate the change in entropy when 1.00 kg of ice is melted at 0 degree c. assume that the heat of fusion of water is lf = 3.34 times 105j/kg. is the change entropy greater for melting or for vaporization? the change entropy greater for melting the change entropy greater for vaporization

Case 1 : Entropy of vaporization of water is

Case 2 :Entropy of fusion of ice is

Result : Entropy of vaporization of water is greater than entropy of fusion of ice

Explanation:

Entropy of closed system is given by

Where is change of entropy, 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 J/kg.

m=1 kg of water

L= J/kg.

Entropy of closed system is given by

Where, T=100+273=373K

Thus, Entropy of vaporization of water is

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 J/kg.

m=1 kg of ice

L=

Entropy of closed system is given by

Where, T=0+273=273K

Thus, Entropy of fusion of ice is

Therefore, Entropy of vaporization of water is greater than entropy of fusion of ice

answer: 802.3 kilo joules of energy represents 191.75 kilo calorie of energy.

explanation:

energy released during combustion of one mole methane = 802.3 kilo joules

if 1 kilo joule = 0.239006   kilo calorie

then ,802.3 kilo joule will be = 802.30.239006 kilo calorie

802.3 kilo joule = 191.75 kilo calorie

hence, the energy in kilo calorie is 191.75 kilo calorie.

less than one you can't create energy