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lara [203]
4 years ago
15

A Carnot engine operates with a cold reservoir at a temperature of 455 K and a hot reservoir at a temperature of 619 K. What is

the net entropy change (in J/K) as it goes through a complete cycle? Round your answer to the nearest whole number.
Physics
1 answer:
Brut [27]4 years ago
4 0

Answer:

Assuming that 4000 J of heat transfer occurs from it the change in entropy will be=2.33j/k

Explanation:

Given data

Th=619 K

Tc= 455K

We are going to assume that 4000 J of heat transfer occurs from it, since it was not specified in the question.

we know that the change in entropy is given as

ΔStotal=ΔShot+ΔScold .

ΔShot=−Q/Th=−4000J/619K=−6.46J/K

For the cold reservoir,

ΔScold=Q/Tc=4000J/455K=8.79J/K

therefore  the total is

ΔStotal=ΔShot+ΔScold

=(−6.46+8.79)J/K

=2.33j/k

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Answer:

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A diffraction grating is to be used to find the wavelength of the emission spectrum of a gas. The grating spacing is not known,
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Answer:

528.9 nm

Explanation:

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