Consider a reversible heat engine that employs a hot reservoir at a temperature of 610 K and a cold reservoir at 290 K. (a) What

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3 years ago

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Consider a reversible heat engine that employs a hot reservoir at a temperature of 610 K and a cold reservoir at 290 K. (a) What

is the entropy change of the hot reservoir during a period in which 4730 J is extracted from the hot reservoir? 7.75 J/K

Physics

1 answer:

Pavlova-9 [17] · Answer 1 · 2022-02-09T04:42:59+03:00

To solve this problem it is necessary that we start from the definition of entropy as a function of heat and temperature exchange. Mathematically this thermodynamic expression can be described as

$\Delta S = \frac{Q}{T}$

Where,

Q= Heat exchange

T = Temperature

Since we look for entropy in the hot reservoir, and considering our given values we have to

$T_H = 610K\\T_C = 290K \\Q_H = 4730J$

Replacing we have:

$\Delta S_H= \frac{Q_H}{T_H}$

$\Delta S_H = \frac{4730J}{610K}$

$\Delta S_H = 7.754J/K$

Therefore the final change in the entropy is 7.75J/K