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garri49 [273]
4 years ago
9

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]4 years ago
6 0

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

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2 years ago
On an aircraft carrier, a jet can be catapulted from 0 to 240 km/h in 2.10 s. If the average force exerted by the catapult is 9.
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<h2>Answer</h2>

The answer is 29495.3 kg

<h2>Explanation</h2>

As we know, the Newton's Second Law of Motion

Force = Mass * Acceleration

F = ma

The F exerted by catapult = 9.35 * 10^5

The Time is = 2.10 s

Velocity of the catapult = 240 km/h = 66.6 m/s   (240*1000 m /3600 s)

The mass of jet = ?

First detrmine the acceleration as

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