# Thermodynamic cycle related problem

1. Oct 9, 2011

### ibysaiyan

1. The problem statement, all variables and given/known data

The problem is as following:

2. Relevant equations

$\Delta$ E = $\Delta$ Q-$\Delta$ W

3. The attempt at a solution
Now I know that if work is done by the system then work sign is negative and that if volume remains unchanged then no work is done.

Am I right to say that Internal energy @ C is 50 joules ?
How does that answer my question ? I know over all energy is conserved in cycle.

Edit i see that the screenshot hasnt loaded i will edir it once i get back home
Thanks!

Last edited: Oct 9, 2011
2. Oct 9, 2011

### ibysaiyan

I have updated the main post.

I think i have worked out solution to this problem.
Given function are:

$\Delta$ Q ABC = 80J.
During this cycle 30j of work is done by the system. Therefore Internal energy at this cyle will be : $\Delta E$ = Q -W = 80-30 = 50j. Ec >EA by 50 joules ?

It's given that work done by the system from A- > D = 20j. Heat entering at cycle DC = 20k aka internal energy. From previous part we know qabc = qab +qbc =70j so we can re arrange this to give us Qbc = 80-10 = 70j.
From A->D 70 joules of heat enters.

For the last answer where it asks for the comparison of internal energies I have got 50joules again.

Can someone please criticize my working and shed some light over this.

Thank you.

Last edited: Oct 9, 2011
3. Oct 9, 2011

### grzz

Part (a) is Ok
Part (b).... why is heat entering during DC be 20k when problem says it is !0?

4. Oct 10, 2011

### ibysaiyan

Sorry I mean't 10j. I don't know why I keep stating in wrong numbers which even the question doesn't mention. So what am I doing wrong for part b ? :/

5. Oct 10, 2011

### grzz

Sorry for my uprupt disappearance last time due a total power failure in our area.

Note that if the transfer of heat energy from D to C is 10J , this DOES NOT mean that the transfer of h.e. from A to B is the same. This is because though the processes DC and AB are both at constant volume, yet they occur under different conditions of P and T.

You may try using the 1st law of thermodynamics for AD and DC.

Last edited: Oct 10, 2011
6. Oct 10, 2011

### ibysaiyan

Hi,
thanks for your reply. As you have pointed out it's senseless of me to assume that heat at DC is similar to AB.
Am I right to say that Qabc - 80j (given) then QADC= 80j too ? [ I get 70 joules as the answer of heat entering into phase AD. Since QADC = AD+ DC =80J ,where DC =10.

What am I missing ?

Last edited: Oct 10, 2011
7. Oct 10, 2011

### grzz

if QABC = 80J then QADC MAY NOT be equal to it because process ABC occurs under different P, V values from process ADC.

But, since the internal energy U of an IDEAL gas depends only on T and the initial and final temperatures of processes ABC and ADC are the same, then the change in INTERNAL ENERGY U is the same for ABC and ADC.

As I said before try using the !st law for AD and for DC.