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 = [tex]\frac{Q}{T}[/tex]
Putting the given values in the above equation as follows.
S = [tex]\frac{Q}{T}[/tex]
= [tex]\frac{546 kJ}{383 K}[/tex]
= -1.4255 kJ/K
As temperature changes from [tex]110 ^{o}C[/tex] to [tex]15 ^{o}C[/tex]. So, (15 + 273) K = 288 K. Hence, change in entropy will be calculated as follows.
[tex]\Delta S = \frac{Q}{T}[/tex]
= [tex]\frac{546 kJ}{288 K}[/tex]
= 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.
