Calculating the waste heat of Carnot engine

[JoeyBob]

Homework Statement

See attached

Relevant Equations

e=1-(Tc/Th)

e=W/Qin

e=(Qin-Qout)/Qout

The answer is 6470 J.

So since I have the two temperatures I could calculate the efficiency. First I convert to kelvin then get an efficiency of 0.35481. Now I can use e=W/Qin to get Qin. I get a value of 10033.54J.

Now I can use e=(Qin-Qout)/Qout to get Qout, the waste heat. I get 7405.9 J.

Overview of calculations:

e=1-(295.66/458.25)=0.35481

Qin=W/e=3560/0.35481=10033.54 J

Qout=Qin/(e+1) = 10033.54/(1.35481) = 7405.9 J but answer is 6470 J.

 

[kuruman]

It looks like you messed up the algebra. I agree with your number for the efficiency but not with one of your expressions for it. For best results, try solving this without numbers, a recommendation that was made to you before.
You have $e=1-\dfrac{Q_{out}}{Q_{in}}$ and $e=\dfrac{W}{Q_{in}}$. Can you eliminate $Q_{in}$ and get an expression relating $Q_{out}$ to $e$ and $W?$

 

[JoeyBob]

It looks like you messed up the algebra. I agree with your number for the efficiency. Try solving this without numbers, a recommendation that you have so far ignored.
You have $e=1-\dfrac{Q_{out}}{Q_{in}}$ and $e=\dfrac{W}{Q_{in}}$. Can you eliminate $Q_{in}$ get an expression relating $Q_{out}$ to $e$ and $W?$

Qin=W/e so

e=1-Qout/(W/e)

e=1-eQout/W

1-e=eQout/W

(1-e)W/e = Qout

(1-0.35481)*3560/0.35481

Youre right, must have messed up my algebra somewhere because this gives the right answer.

Thanks.

 

[kuruman]

Youre right, must have messed up my algebra somewhere because this gives the right answer.

It's not the algebra, it's this: e=(Qin-Qout)/Qout.

When Qout = 0 you should expect efficiency of 1. Your expresion does not predict that. That is why I prefer to write it as $e=1-\dfrac{Q_{out}}{Q_{in}}.$ It reduces to 1 when no heat is wasted. I am mentioning this so that you will write the correct expression next time.

 

[Chestermiller]

I would never go to the efficiency to solve this. I would write $$Q_{in}-Q_{out}=W$$ and $$\frac{Q_{in}}{T_{hot}}-\frac{Q_{out}}{T_{cold}}=0$$This is two linear algebraic equations in two unknowns.