91Ó°ÊÓ

We know that the internal energy of the system doesn’t change for a cyclic process.

i e.$∆U=0$

In a cyclic process the clockwise process gives positive work and the anti-clock process gives the negative work.

$$\begin{gathered} W_{cw}>0 \\ {W}_{acw}<0 \end{gathered}$$

Now, from the figure it is clear that,

• Loop l is clockwise cycle, therefore the work done is positive.
• Loop ll is anti-clock wise, therefore the work done is negative.
• Loop l is larger than loop

Since, Loop l is larger than Loop ll , the total work done in a full cycle is positive, I,e.

$W_t>0$

Since the internal energy is zero and the total work in a full cycle is positive I,e.$W_t>0$ .

Therefore, according to first law of thermodynamics we have

$$\begin{gathered} ∆Q=∆U+W \\ ∆Q=0+Wt \\ ∆Q=Wt \\ ∆Q>0 \end{gathered}$$

Hence, over one complete cycle the heat $∆Q$is positive. Therefore, the heat flows into the system

In a cyclic loop we have$∆Q=W_t$

For Loop l

Since the work done in Loop l is positive and the change in internal energy is zero.

Therefore, from the first law of thermodynamics we have:

$$\begin{gathered} ∆Q=W_l \\ ∆Q>0 \end{gathered}$$

Hence, for the loop l the heat $∆Q$is positive and it flows into the system.

For Loop ll

The work done in Loop ll is negative i.e,$W_{ll}>0$ , and the change in internal energy is zero. Therefore, from the first law of thermodynamics we have:

$$\begin{gathered} ∆Q=W_{ll} \\ ∆Q<0 \end{gathered}$$

Therefore, in Loop ll the heat$∆Q$ is negative and the heat flows out of the system.