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One end of a metal rod is in contact with a thermal reservoir at 699. K, and the other end is in contact with a thermal reservoir at 101. K. The rod and reservoirs make up a closed system. If 6970. J are conducted from one end of the rod to the other uniformly (no change in temperature along the rod) what is the change in entropy of a) each reservoir, b) the rod and c) the system?

Respuesta :

Answer:

a)ΔS₁ = - 9.9 J/K

ΔS₂ = 69 J/K

b)The entropy change for the rod = 0 J/K

c)ΔS = 59.1 J/K

Explanation:

Given that

T₁ = 699 K

T₂= 101 K

Q= 6970 J

Change in entropy given as

[tex]\Delta S=\dfrac{Q}{T}[/tex]

For 699 K:

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

[tex]\Delta S_1=-\dfrac{6970}{699}[/tex]

ΔS₁ = - 9.9 J/K  ( Negative because heat is leaving from the system)

For 101 K;

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

[tex]\Delta S_2=\dfrac{6970}{101}[/tex]

ΔS₂ = 69 J/K

The entropy change for the rod = 0 J/K

Entropy  change for the system

ΔS = ΔS₂  + ΔS₁

ΔS = 69 -9.9 J/K

ΔS = 59.1 J/K

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