Corrected.Finding the Energy and Work in a Carnot Cycle

[Feodalherren]

Homework Statement

An ideal gas is taken through a Carnot cycle. The
isothermal expansion occurs at 250°C, and the isothermal
compression takes place at 50.0°C. The gas takes in 1 200 J
of energy from the hot reservoir during the isothermal
expansion. Find (a) the energy expelled to the cold reservoir
in each cycle and (b) the net work done by the gas in
each cycle.

Homework Equations

The Attempt at a Solution

Qh=1200J.

Qh-W=Qc (absolute values)

e(Carnot)=(523-323)/523= 61.9 %
(corrected my typo from before)

Therefore Qc= .381(1200J)=457J

and W = 1200-457=743J

Not the correct answer. Qc = 741 according to the solutions manual.

ps. Latex seems to be bugged, didn't work in "preview post" mode.

 

[TSny]

e(Carnot)=1-(523/323)= 61.9 %

Hello. Check this calculation.

 

[Feodalherren]

Hi there.

Woops I typed it in wrong, it was supposed to be (523-323)/523= .619

The answer is still incorrect though.

 

[SteamKing]

Hi there.

Woops I typed it in wrong, it was supposed to be (523-323)/523= .619

The answer is still incorrect though.

That's not the problem. Check the numbers. Remember, you are looking for a difference of 2 joules.

 

[Feodalherren]

Give me another hint, please. I still don't know what I'm doing wrong :). Is my logic even sound?

 

[TSny]

Hi there.

Woops I typed it in wrong, it was supposed to be (523-323)/523= .619

The answer is still incorrect though.

e = (523-323)/523 is correct. But this does not equal .619

 

[Feodalherren]

what?! How did I miss that, I re-did it like 3 times :/. Thank you though.

 

[TSny]

Hi there.

Woops I typed it in wrong, it was supposed to be (523-323)/523= .619

The answer is still incorrect though.

Also, to get the anwer of 741 J it appears that you should convert to Kelvin by adding 273.15 rather than 273.