Question

One end of a copper rod is in thermal contact with a

Answer 1

Answer:

$\Delta S_1=-16\ J/K$

$\Delta S_2=26.66\ J/K$

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 :

$\Delta S_1=-\dfrac{Q}{T_1}$

$\Delta S_1=-\dfrac{8000}{500}$

$\Delta S_1=-16\ J/K$

The entropy change for the cold  reservoir :

$\Delta S_2=\dfrac{Q}{T_2}$

$\Delta S_2=\dfrac{8000}{300}$

$\Delta S_2=26.66\ J/K$

The entropy for universe

ΔS=ΔS₁+ΔS₂

ΔS= - 16 + 26.66 J/K

ΔS= 10.66 J/K