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

0.64

Step-by-step 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

`(T_(c))/(T_(H))=(Q_(c))/(Q_(H))`

`(T_(c))/(T_(H))=(290)/(450)=0.64`

So, the ratio is 0.64.

Answer 2

So the ratio will be
`(T_L)/(T_H)=-0.171`

Step-by-step 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 =(W)/(Q)=(290)/(450)=0.6444`

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

So efficiency of Carnot engine
`=(0.6444)/(0.55)=1.171`

Efficiency of Carnot engine is
`\eta =1-(T_L)/(T_H)`

`1.171 =1-(T_L)/(T_H)`

`(T_L)/(T_H)=-0.171`