Question

A heat engine runs in the cycle A to B to C and back to A as shown in the PV diagram. (See Figure 1 below.) The pressure at C is 200 kPa. The volume at C is 16 liters. The temperature at C is T K. The pressure at A is twice that at C. The volume at B is twice that at C. The internal energy at A and B are both 9600 J. The internal energy at C is 4800 J.

***THE TEMPERATURE POINT C IS 300 K!***

Answer 1

a) What is the work done in the transition A to B?
Express your answer using three significant figures in unit of J.

b) What is the work done in the transition B to C? Answer is in unit J.

c) What is the TOTAL heat going into the gas during the cycle? Answer is in unit J.

d) What is the total work done during the cycle? Answer is in unit J.

e) What is the efficiency of this heat engine? Answer is in %.

f) What is the efficiency of a Carnot cycle running between the lowest and highest temperatures of the cycle ABC? Answer is in %.

Answer 2

`a) The work done going from A to B is the area under the curve: (1/2)(16 liters)(200 kPa)+200 kPa * 16 liters. This is the area of the triangle and the imaginary rectangle below it. You will need to convert liters to cubic meters and use Pa rather than kPa so that work comes out in SI units (Joules). 1,000 liters = 1 m^3`

`b) The work in going from B to C is the area of the PV diagram of the imaginary rectangle between V and 2V an 0 and 200 kPa. (convert to SI)`

`c) The Total heat going into the gas is Qin = 9600-4800 J, the difference in internal heat`

`d) The amount of work done is the difference in area between parts b and c above: Work = Area from A to B - Area from B to C`

`e) The efficiency is 1 - Work/Qin (see parts c and d above for Work and Qin)`

`f) Carnot Eff = 1 - Tc/Th , where Tc=294 K, temp of ambient environment and Th = 300 K. You'll then need to convert to %`

Answer 3

I tried a) and got:
(1/2)(0.016 m^3)(200000 Pa) + (200000 Pa)(0.016 m^3)
= 1600 + 20000.016
= 201600.016 J
>>> 3 sig figs = 202000 J

The program is saying that is wrong. What did I mess up?

Answer 4

I skipped b) because it's a similar concept.

c) 9600J - 4800J = 4800 J ; It said this was wrong too.

I skipped d) and e) because I can't get this right if I don't have the other ones right.

e) 1 - 294 K/300 K
= 1 - 0.98
= 0.02 * 100 = 2% ; It said this was wrong too.

Answer 5

I don't see the problem, maybe someone else will see the issue.

Answer 6

My teacher said that we didn't have to do this one today.
I have no idea if it was due to how the program set it up or not?
Thanks anyway.