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.