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. Assum
1 answer:
Answer:
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
You might be interested in
(a) first two overtones for each string:
The first string has a fundamental frequency of 196 Hz. The n-th overtone corresponds to the (n+1)-th harmonic, which can be found by using
where f1 is the fundamental frequency.
So, the first overtone (2nd harmonic) of the string is
while the second overtone (3rd harmonic) is
Similarly, for the second string with fundamental frequency
, the first overtone is
and the second overtone is
(b) The fundamental frequency of a string is given by
where L is the string length, T the tension, and
is the mass per unit of length. This part of the problem says that the tension T and the length L of the string are the same, while the masses are different (let's calle them 
, the mass of the string of frequency 196 Hz, and 
, the mass of the string of frequency 523 Hz.
The ratio between the fundamental frequencies of the two strings is therefore:
and since L and T simplify in the equation, we can find the ratio between the two masses:
(c) Now the tension T and the mass per unit of length
is the same for the strings, while the lengths are different (let's call them 
and 
). Let's write again the ratio between the two fundamental frequencies
And since T and simplify, we get the ratio between the two lengths:
(d) Now the masses m and the lenghts L are the same, while the tensions are different (let's call them
and 
. Let's write again the ratio of the frequencies:
Now m and L simplify, and we get the ratio between the two tensions:
The first string has a fundamental frequency of 196 Hz. The n-th overtone corresponds to the (n+1)-th harmonic, which can be found by using
where f1 is the fundamental frequency.
So, the first overtone (2nd harmonic) of the string is
while the second overtone (3rd harmonic) is
Similarly, for the second string with fundamental frequency
and the second overtone is
(b) The fundamental frequency of a string is given by
where L is the string length, T the tension, and
The ratio between the fundamental frequencies of the two strings is therefore:
and since L and T simplify in the equation, we can find the ratio between the two masses:
(c) Now the tension T and the mass per unit of length
And since T and
(d) Now the masses m and the lenghts L are the same, while the tensions are different (let's call them
Now m and L simplify, and we get the ratio between the two tensions: