Question

3. In a Carnot's cycle, the highest temperature is ntimes the lowest temperature. Also the heat givento the system during the cycle is n times total workdone during the cycle. What is the value of n ?​

Answer 1

Answer:

The ideal Rankine cycle efficiency (or Carnot cycle efficiency) can be defined as (T2-T1)/T2. T2 is defined as the absolute temperature of the heat source and T1 is the absolute temperature of the heat sink.

Answer 2

The value of n is 2.

Given:

1. Highest temperature is n times the lowest temperature.
2. Heat given is n times the work done.

To find:

The value of n.

Solution:

Let us assume that the Carnot engine works between the source temperature $T_{1}$ and the sink temperature $T_{2}$. The source is at a higher temperature hence, $T_{1}$ > $T_{2}$ .

The work done by the system is $W$ and $Q$ is the heat given to the system. Therefore,

According to the question,

$T_{1}=n T_{2}$                        $(i)$

$Q=nW$                         $(ii)$

We know,

The efficiency $(C)$ of the Carnot engine is given by,

$C=\frac{W}{Q} =1-\frac{T_{2} }{T_{1} }$  

Substituting the known values in the equation, we get

$\frac{W}{Q} =1-\frac{T_{2} }{T_{1} }$

$\frac{W}{nW}=1- \frac{T_{2} }{nT_{2} }$

$\frac{1}{n}=1- \frac{1}{n}$

$\frac{2}{n}=1$

$n=2$

Final answer:

Hence, the value of n is 2.