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One end of a copper rod is in thermal contact with a
hotreservoir at T=500K, and the other end is in thermal contact
with acooler reservoir at T=300K. If 8000 J of energy is
transferred fromone end of the rod to the other, with no change in
the temperaturedistribution, find the entropy change of each
reservoir and thetotal entropy change of the Universe.

Answer :

Answer:

[tex]\Delta S_1=-16\ J/K[/tex]

[tex]\Delta S_2=26.66\ J/K[/tex]

ΔS= 10.66 J/K

Explanation:

Given that

For hot reservoir :

T₁= 500 K

For cold reservoir :

T₂= 300 K

Energy transfer ,Q= 8000 J

The entropy change for the Hot reservoir :

[tex]\Delta S_1=-\dfrac{Q}{T_1}[/tex]

[tex]\Delta S_1=-\dfrac{8000}{500}[/tex]

[tex]\Delta S_1=-16\ J/K[/tex]

The entropy change for the cold  reservoir :

[tex]\Delta S_2=\dfrac{Q}{T_2}[/tex]

[tex]\Delta S_2=\dfrac{8000}{300}[/tex]

[tex]\Delta S_2=26.66\ J/K[/tex]

The entropy for universe

ΔS=ΔS₁+ΔS₂

ΔS= - 16 + 26.66 J/K

ΔS= 10.66 J/K

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