Question

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

Answer 1

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 $\eta =1-\frac{T_L}{T_H}$

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

$\frac{T_L}{T_H}=-0.171$

Answer 2

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.