Thermodynamic cycle related problem

[ibysaiyan]

Homework Statement

The problem is as following:

Homework Equations

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

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.

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

 

[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 ?

Now to answer part b)
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.

 

[grzz]

Now to answer part b)
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.

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

 

[ibysaiyan]

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

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 ? :/

 

[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.

 

[ibysaiyan]

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.

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 ?

 

[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.