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ycow [4]
3 years ago
5

One end of a metal rod is in contact with a thermal reservoir at 745. K, and the other end is in contact with a thermal reservoi

r at 101. K. The rod and reservoirs make up a closed system. Of 7190. 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 reservoirb. the rodc. the system
Physics
1 answer:
Masteriza [31]3 years ago
3 0

Answer:

a)S_1=-9.65}\ J/K

b)S_2=71.18\ J/K

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

S_1=-\dfrac{7190}{745}\ J/K

Negative sign because heat is leaving.

S_1=-9.65}\ J/K

b)

The entropy change of reservoir 101 K

S_2=\dfrac{7190}{101}\ J/K

S_2=71.18\ J/K

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

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