A frictionless piston-cylinder device contains a saturated liquid and water vapor mixture at 110 ℃. During a constant pressure p
1 answer:
Explanation:
(a) It is given that heat transferred is -546 kJ (as heat is releasing) and temperature is 110 degree celsius.
Hence, the temperature in kelvin will be as follows.
(110 + 273) K
= 383 K
As it is known that relation between entropy change and heat is as follows.
S =
Putting the given values in the above equation as follows.
S =
=
= -1.4255 kJ/K
As temperature changes from to
. So, (15 + 273) K = 288 K. Hence, change in entropy will be calculated as follows.
=
= 1.895 kJ/K
Therefore, entropy change of water will be 1.895 kJ/K.
(b) As total entropy generated will be the sum of both the entropies as follows.
Total entropy = (-1.4255 kJ/K + 1.895 kJ/K)
= 0.4695 kJ/K
Thus, total entropy during this heat transfer process is 0.4695 kJ/K.
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