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Thermodynamics, solving for Q1. I've almost got it!

  • #1
208
7

Homework Statement



A reversible engine takes in 2000J from a reservoir at 400K; it gives up Q1' J to a reservoir at 300K and Q1 J to a reservoir at 200K, doing 750J of work/cycle. Find Q1' and Q1.

Homework Equations



So at first glance the relevant equation would seem to be the work equation W=Q2-Q1, however this wouldn't account for the difference in the two reservoir temperatures. So my next thought was to use the proportions Q1/Q2 = T1/T2.

The Attempt at a Solution



If I go with my gut and use the proportion what I get is:
Q1/Q2 = T1/T2
Q1/2000 = 300/400
Q1=1500J

and

Q1/2000 = 200/400
Q1 = 1000J

Now, I can tell almost by inspection that these results are way off (and indeed the book values give are 750J and 500J). If I attempt using the basic W = Q2-Q1, I get W = 1250, but again this doesn't account for temperature (but is conveniently the sum of the given answers).

What am I doing wrong? Any nudge in the right direction would be appreciated.
 

Answers and Replies

  • #2
3,003
2
In general the approach to these problems is to apply the energy balance (1st law) and entropy balance (2nd law) together.

So from the energy balance: [itex]\Delta E_{engine} = 0 = Q1 - Q1' - W \Rightarrow Q1 - Q1' = W[/itex]

From the entropy balance: [itex]\Delta S_{engine} = 0 = Q1/T1 - Q1'/T1' \Rightarrow Q1/T1 = Q1'/T1'[/itex]

So you are almost there! You need to use both equations to take the temperatures into account (i.e., solve the entropy balance for one of the Q's and plug it into energy balance).
 
  • #3
208
7
Ok, then does the following make sense (ignoring the fact that I need to learn latex way better):

Q1 = WT1/T2-T1

Q1' = (T2/T1)Q1
 
  • #4
208
7
No, my second post doesn't work out. Always good to know you're close though!
 
  • #5
208
7
Ok, pulling my head out of nonsense and actually doing basic maths I get the following:

W = (Q1'/T1')T - Q1'
 
  • #6
208
7
Got it! Thanks!
 
  • #7
3,003
2
Yay! :smile:
 

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