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

the answer is c. the lower legs are levers, and the knees are fulcrums. the ankles hold the loads.

96.77 amp

explanation:

rl = 250/100 = 2.5 ohms.

egl = 250 + (100 × 0.1) + 2 = 262 v.

at 700 r.p.m., eg2 = 262 × 700/1000 = 183.4 v.

if ia is the new current, then eg2 - 2 - (ia × 0.1) = 2.5 ia

which means this will give ia = 96.77 amp.