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sertanlavr [38]
3 years ago
8

In one cycle a heat engine absorbs 450 J from a high-temperature reservoir and expels 290 J to a low-temperature reservoir. If t

he efficiency of this engine is 55% of the efficiency of a Carnot engine, what is the ratio of the low temperature to the high temperature in the Carnot engine?
Physics
2 answers:
makvit [3.9K]3 years ago
4 0

Answer:

0.64

Explanation:

QH = 450 J

Qc = 290 J

efficiency, η = 55 %

Let Tc be the low temperature and TH be the high temperature.

According to the Carnot's theorem

\frac{T_{c}}{T_{H}}=\frac{Q_{c}}{Q_{H}}

\frac{T_{c}}{T_{H}}=\frac{290}{450}=0.64

So, the ratio is 0.64.

Vitek1552 [10]3 years ago
3 0

Answer:

So the ratio will be \frac{T_L}{T_H}=-0.171

Explanation:

We have given heat engine absorbs 450 joule from high temperature reservoir

So Q=450j

As the heat engine expels 290 j

So work done W = 290 J

We know that efficiency \eta =\frac{W}{Q}=\frac{290}{450}=0.6444

It is given that efficiency of the engine only 55 % of Carnot engine

So efficiency of Carnot engine =\frac{0.6444}{0.55}=1.171

Efficiency of Carnot engine is \eta =1-\frac{T_L}{T_H}

1.171 =1-\frac{T_L}{T_H}

\frac{T_L}{T_H}=-0.171

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Part B) the angular velocity is 195.13 rad/s

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