Though the two statements appear to be different, but they are equivalent, they mean the same thing. This is verified as follows.

**6.6.1 Violation of Clausius Statement Leads to Violation of Kelvin-Plank Statement**

Let us take a refrigerator ‘R’ which violates Clausius statement requiring no work to transfer heat Q_{2} from low temperature T_{2} to high temperature T_{1}.

Now we take a heat engine (E) which follows Kelvin-Plank statement i.e., it draws heat Q_{1} from T1 and rejects heat Q_{2 }to T_{2}. Thereby it produces work W which is equal to Q_{1} – Q_{2} (Ref Fig. 6.11 (a)).

Now let us couple the two engines and consider a combined system as shown in Fig. 6.11 (b). The combined system draws heat from T _{1} which is equal to Q_{1} – Q_{2} and converts completely to equal amount of work W = Q_{1} – Q_{2} without rejecting any heat to lower temperature body. This violates Kelvin-Plank statement.

**6.6.2 Violation of Kelvin-Plank Statement Leads to Violation of Clausius Statement**

Fig. 6.12 (a) shows a heat engine E which violates Kelvin-Plank statement i.e., it draws

heat of amount Q_{1 }– Q_{2 }and converts completely to work W = Q_{1} – Q_{2}. Let us take a refrigerator R which follows Clausius statement i.e., it draws heat Q2 from low temperature T_{2 }and inject heat Q_{1} to high temperature T1 and it requires work W = Q_{1} – Q_{2}.

If we couple R with E, the work from E is directly given to R. The combined system is shown in Fig. 6.12 (b). Fig. 6.12 (b) shows that the combined system draws heat of amount Q_{2 }from low temperature T_{2} and inject same amount Q_{2} to higher temperature T_{1} without any work done on it. Thus this violates Clausius statement

**SOLVED EXAMPLES**

**Example 6.1.** An engine inventor claims to have developed an engine that takes in 140 MJ at

a temperature of 405 K and rejects 45 MJ at a temperature of 205 K while producing 20 KWh

of mechanical work. Find if the claim of inventor is true or false.

**Solution.** Q_{1} = J 40 MJ, Q_{2 }= 45 MJ, W = 20 KWh

W = 20 KWh= 20x 3600 KJ = 72 MJ.

W + Q_{2} = (72 + 45) MJ = 117 MJ

Q_{1} = 140MJ

Thus Q_{1} = W + Q_{2} àViolating 1st law of thermodynamics (energy not conserved)

(Th) _{max} = 1 – T_{2} / T_{1} = 1 – 205 / 405 = 0.49

( Th) _{claimed} = W / Q_{1 }= 72 / 140 = 0.51

Thus (Th)claimed > (Th)max à It violates the 2nd law also.

Thus the claim is false.

**Example 6.2.** If a refrigerator is sued for heating purposes in winter so that the atmosphere becomes the cold body and the room to be heated becomes the hot body, how much heat would be available for heating for every 1.5 KW input to the driving motor? The COP of the refrigerator is 6, and the electromechanical efficiency of the motor is 80%.

**Solution.**

Here motor efficiency= 0.8

Motor output / input = 0.8

Motor output / 1.5 = 0.8

Motor output= 0.8 x 1.5 KW

= W to Refrigerator

COP net = Q_{2} / W

6 = Q_{2} / 0.8 x 1.5 or Q_{2} = 7.2 KW

Q_{1} = W + Q_{2} = 0.8 x 1.5 + 7.2

= 8.4 kw

Heat available = 8.4 kw.

**Example 6.3.** An electric storage battery which can exchange heat only with a constant

temperature atmosphere goes through a complete cycle of two processes. In process 1-2, 2.8 KWh of electrical work flow into the battery while 732 KJ of heat flow out to the atmosphere. During process 2 – 1, 2.4 KWh of work flow out of the batten;. (a) Find the heat transfer in process 2 – 1 (b) What is the maximum possible work of 2 – 1 process if 1 – 2 process has occured as above? (c) What will be the heat transfer in the condition of maximum possible work output in process 2 – 1?

** **

**Solution.** For process 1 – 2

Q_{1-2}= U_{2}-U_{1} + W_{l-2}

– 732 = u _{2} – U_{1} – 2.8 x 36oo

U_{2} – U; = 9348 KJ

For process 2 – 1

Q_{2-1} = ul – U_{2} + w_{2-1}

= – 9348 + 2.4 X 3600

= – 708 KJ Ans. (a)

(Negative sign shows that heat flow out of the system)

Q_{2-1 }= U_{1 }– U_{2} + w_{2-l}

For maximum work Q_{2-1} = 0

W_{2-1})max = U_{2}-U_{1} = 9348KJ. Ans. (b)

Q2-1 = 0 for maximum work. Ans. (c)