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Dovator [93]
3 years ago
7

Heat is transferred at a rate of 2 kW from a hot reservoir at 825 K to a cold reservoir at 300 K. Calculate the rate at which th

e entropy of the two reservoirs changes. (Round the final answer to six decimal places.) The rate at which the entropy of the two reservoirs changes is 00424 kW/K.
Chemistry
1 answer:
kakasveta [241]3 years ago
6 0

Answer:

0.004243 KW/K.

Explanation:

From the question given above, the following data were obtained:

Heat transfered (Q) = 2 KW

Temperature of hot reservoir (Tₕ) = 825 K

Temperature of cold reservoir (T꜀) = 300 K

Next, we shall determine the entropy of both reservoir. This can be obtained as follow:

For hot reservoir:

Heat transfered (Q) = 2 KW

Temperature of hot reservoir (Tₕ) = 825 K

Entropy of hot reservoir (Sₕ) =?

Sₕ = Q/Tₕ

Sₕ = 2/825

Sₕ = 0.002424 KW/K

For cold reservoir:

Heat transfered (Q) = 2 KW

Temperature of cold reservoir (T꜀) = 300 K

Entropy of cold reservoir (S꜀) =?

S꜀ = Q/T꜀

S꜀ = 2/300

S꜀ = 0.006667 KW/K

Finally, we shall determine the change in the entropy of the two reservoirs. This can be obtained as follow:

Entropy of hot reservoir (Sₕ) = 0.002424 KW/K

Entropy of cold reservoir (S꜀) = 0.006667 KW/K

Change in entropy (ΔS) =?

ΔS = S꜀ – Sₕ

ΔS = 0.006667 – 0.002424

ΔS = 0.004243 KW/K.

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