Answer:
a)[tex]S_1=-9.65}\ J/K[/tex]
b)[tex]S_2=71.18\ J/K[/tex]
c) 0 J/K
d)S= 61.53 J/K
Explanation:
Given that
T₁ = 745 K
T₂ = 101 K
Q= 7190 J
a)
The entropy change of reservoir 745 K
[tex]S_1=-\dfrac{7190}{745}\ J/K[/tex]
Negative sign because heat is leaving.
[tex]S_1=-9.65}\ J/K[/tex]
b)
The entropy change of reservoir 101 K
[tex]S_2=\dfrac{7190}{101}\ J/K[/tex]
[tex]S_2=71.18\ J/K[/tex]
c)
The entropy change of the rod will be zero.
d)
The entropy change of the system
S= S₁ + S₂
S = 71.18 - 9.65 J/K
S= 61.53 J/K
