Heat Q flows spontaneously from a reservoir at 383 K into a reservoir

[copitlory8]

Heat Q flows spontaneously from a reservoir at 383 K into a reservoir that has a lower temperature T. Because of the spontaneous flow, thirty percent of Q is rendered unavailable for work when a Carnot engine operates between the reservoir at temperature T and a reservoir at 234 K. Find the temperature T.

I have absolutely no idea. Please help me. I am getting desperate. I have to pull up my failing physics grade or else i will pretty much be disowned by my family.

 

[rock.freak667]

Because of the spontaneous flow, thirty percent of Q is rendered unavailable for work

Given this, what would the efficiency of the Carnot Engine?

When you get that, what is the expression for efficiency of a Carnot Engine?

 

[copitlory8]

the efficiency i guess is 2/3.
e=1-(Tc/Th)

 

[copitlory8]

so what next?

 

[rock.freak667]

the efficiency i guess is 2/3.

How did you get 2/3 ?

e=1-(Tc/Th)

So when you get 'e' T_h=T, and you were given T_c, can you get T?

 

[copitlory8]

i got 2/3 because it says that 1/3 is unavailable therefore 2/3 is available

 

[rock.freak667]

i got 2/3 because it says that 1/3 is unavailable therefore 2/3 is available

That thinking is correct but thirty percent is not the same as 1/3.

 

[copitlory8]

oh no! I am lost again what is the th?

 

[copitlory8]

so the efficiency is actually .7

 

[rock.freak667]

so the efficiency is actually .7

Right so, what is T?

 

[copitlory8]

there are two Ts in the problem. do i use the higher one for Th?

 

[rock.freak667]

there are two Ts in the problem. do i use the higher one for Th?

You are only considering the two temperatures associated with the Carnot engine, the high temperature T and the lower temperature 234 K.

 

[copitlory8]

so i use 234K?

 

[rock.freak667]

so i use 234K?

Yes.

 

[copitlory8]

so let me get this straight .7(234)=T where T is the answer I am looking for? this seems too simple

 

[rock.freak667]

so let me get this straight .7(234)=T where T is the answer I am looking for? this seems too simple

e=1- (T_c/T), so you '0.7' would be '0.3'.

Nothing is being done to the heat when it is moving from 384K to T, so I don't think you'd need to worry about that for now.

 

[copitlory8]

so .3=1-(234/t)

 

[rock.freak667]

so .3=1-(234/t)

No, 0.7 = 1 -(T_c/T)

when you rearrange you'd get the '0.3'.

 

[copitlory8]

alright i think i almost got this. so the T i am trying to solve for is equal to Tc/.3

 

[rock.freak667]

alright i think i almost got this. so the T i am trying to solve for is equal to Tc/.3

Yes and you know th vaue of T_c.

 

[copitlory8]

then what is the number 383 K for?

 

[rock.freak667]

then what is the number 383 K for?

Not too sure why it was given in the problem.

 

[copitlory8]

i did 234/.3=780 and got the wrong answer.

 

[rock.freak667]

Yes I realized, I've asked the other HW helpers to see if they can clarify my error in thinking.

 

[copitlory8]

thanks

 

[diazona]

Heat Q flows spontaneously from a reservoir at 383 K into a reservoir that has a lower temperature T. Because of the spontaneous flow, thirty percent of Q is rendered unavailable for work when a Carnot engine operates between the reservoir at temperature T and a reservoir at 234 K. Find the temperature T.

Odd problem

Anyway, here is my understanding of the problem. A certain amount of thermal energy Q flows from the 383 K reservoir into the reservoir at T. Then a certain different amount of thermal energy (let's call it Q') is transferred from that reservoir, through a Carnot engine, to a reservoir at 234 K. Hopefully you can agree that it makes sense that Q' < Q, although I'm not sure if/how it can be proven that this condition must hold, and that 234 K < T < 383 K.

Now, the problem says that 30% of Q is rendered unavailable for work in the Carnot engine. Here is where it gets confusing. One way to read that is that the spontaneous flow brings the total amount of energy unavailable for doing work up to 30% of Q. This implies that the remaining 70% of Q is available and is used for doing work in the Carnot engine - basically that W = 0.7 Q. So you have W for the heat engine, in relation to Q, but not in relation to Q', which is the actual amount of heat taken in by the engine. To figure out the efficiency of the engine (and thus the temperature of its hot reservoir), you would have to relate Q to Q'.

The other possible reading I see is that 30% of the thermal energy Q is removed from the flow before the energy ever reaches the engine. This would be saying that Q - Q' = 0.3 Q. But with this reading, you have no information about the operation of the engine itself - you don't know anything about its work output or its efficiency. I don't really see where you would go from there.

 

[copitlory8]

so i should just give up and fail

 

[diazona]

so i should just give up and fail

Why would you do that?

 

[copitlory8]

i should just give up because this problem is so ridiculous, not even my awesome physics teacher can solve it after i spent 45 minutes with him on it.

 

[diazona]

It seems quite unreasonable for your teacher to assign you and expect you to solve a homework problem which he himself can't do. But we're not here to tell you what to do, only to help you understand the physics.

 

[copitlory8]

well i understand but my teacher is really smart. and the way webassign works is that is has a huge databank of questions released from my textbook publisher and it picks questions randomly. so my teacher hasn't even seen all the questions so once in a while a tough one appears such as this one. can someone seriously just tell me the solution

 

[diazona]

No, we will not just give you the solution. If you think you deserve to be just given the answer, ask your teacher. Instructors who use WebAssign have access to a complete answer key, and in some cases explanations of how the problems are solved.

As long as you are willing to continue to make an effort on the problem, we can do our best to guide you through it.