Answer:2.517 J/K
Explanation:
Given
Reservoir 1 Temperature [tex]T_1=781 K[/tex]
Reservoir 2 Temperature [tex]T_2=335 K[/tex]
Let Q is the amount of heat Flows i.e. [tex]Q=1477 J[/tex]
thus change in Entropy is given by [tex]\frac{\sum Q}{T}[/tex]
[tex]\Delta S=\frac{\sum Q}{T}=-\frac{Q}{T_1}+\frac{Q}{T_2}[/tex]
[tex]\Delta S=\frac{\sum Q}{T}=-\frac{1477}{781}+\frac{1477}{335}[/tex]
[tex]\Delta S=\frac{\sum Q}{T}=-1.891+4.4089[/tex]
[tex]\Delta S=\frac{\sum Q}{T}=2.517 J/K[/tex]
