Question

A closed system undergoes an adiabatic process during which the work transfer into the system is 12 kJ. The system then returns to its original state via a non-adiabatic process during which the work transfer out of the system is 5 kJ. What is the magnitude and direction of the heat transfer for the second process?

Answer 1

Q= - 7 KJ

Heat is rejected

Step-by-step explanation:

In first process:

Process is adiabatic so heat transfer will be zero.

Q= 0 For adiabatic process

Now from first law of thermodynamics

Q = ΔU + W

ΔU is the change in internal energy

Q is the heat transfer

W is the work.

So here Q= 0

And work is transfer to the system .It means that work done on the system so it will be taken as negative.

W= - 12 KJ

Q = ΔU + W

0 = ΔU -12

ΔU = 12 KJ

It means that in first process internal energy of system increase.

In second process:

non-adiabatic process and work done by system is 5 KJ.

Q = ΔU + W

U is the point function and it does not depends on the path follows it depends only on final and initial states.

So fro second process

ΔU = - 12 KJ

Q= -12 + 5

Q= - 7 KJ

In magnitude Q= 7 KJ

It mean that heat is rejected from the system because it give negative sign.