Respuesta :
Explanation:
The given data is as follows.
Q = 2 kW, [tex]T_{1}[/tex] = 775 K
[tex]T_{2}[/tex] = 300 K
The relation between entropy and heat energy is as follows.
[tex]\Delta S = \frac{Q}{\Delta T}[/tex]
Therefore, calculate the entropy at each temperature as follows.
[tex]S_{1} = \frac{Q}{T_{1}}[/tex]
= [tex]\frac{2 kW}{775 K}[/tex]
= [tex]2.5 \times 10^{-3}[/tex] kW/K
Also, [tex]S_{2} = \frac{Q}{T_{2}}[/tex]
= [tex]\frac{2 kW}{300 K}[/tex]
= [tex]6.6 \times 10^{-3}[/tex] kW/K
Hence, the change in entropy will be calculated as follows.
[tex]\Delta S = S_{2} - S_{1}[/tex]
= [tex](6.6 \times 10^{-3} - 2.5 \times 10^{-3})[/tex] kW/K
= [tex]4.1 \times 10^{-3}[/tex] kW/K
or, = 0.0041 kW/K
Thus, we can conclude that the rate at which the entropy of the two reservoirs changes is 0.0041 kW/K.
