Important questions in Thermodynamics.

Solutions for imp Question of Termal. 

Q1)

 Solution:
 Let heat added is=Q
Then, heat rejected is=0.8×Q
So, net work done(W) =Qadd-Qrej
                            W =Q-0.8×Q
                            W =0.2Q
And efficiency =(net work/heat added)
                            =0.2Q/Q
                            =20%
efficiency of heat engine =                                                                       [1/(1+COPref)]
ie  0.2=[1/(1+COPref)]
        COP ref=(1/0.2)-1
        COP ref=4  (answer)


Q2)

Solution:
For carnot cycle
Sum of all [heat transfer/temperature]                                                                         =0
ie.

(Qadd/T1)+(Qrej/T2) =0
(50/1300)+(Qrej/400) =0
Qrej = -(50/13)×4
Qrej = - 15.38 kW(answer )(- sign represents heat flowing from system)

Q3)

Q4)

Solution:
Internal energy U= m×cp×dT
dT=is degree of molecular activity

Cp=depends only on molecular structure nad characteristic gas constent.

 m= n×M(M is molecular weight )

So answer is D( 1, 2&3).