Thermodynamics Problem: Calculate the heat input for this work output

  • #1

Homework Statement


An inventor claims to have developed a device with the following properties:
net work output per cycle = 20,000 J
heat output per cycle = 50,000 J
high temperature = 450 C
low temperature = 160 C

Calculate the heat input per cycle.

Homework Equations


ΔETH = W + Q
efficiency max = 1 - (Tc/Th)

The Attempt at a Solution


I tried taking efficiency max, which I got 0.40111, and dividing the heat output by that to get 124,654 J. I'm not sure if that's what I'm supposed to be doing, still confused on this.
 
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Answers and Replies

  • #2
kuruman
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The inventor's engine does not necessarily operate at maximum efficiency. Can you find the actual efficiency of this engine? You will need to find the heat input per cycle first. Hint: Use the First Law over a complete cycle.
 
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  • #3
Sorry for the late reply. I know I need to find the heat input per cycle, that's the homework question. By first law do you mean ΔU = Q + W?
 
  • #6
kuruman
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Is it the difference in heat/temperature?
I think @Chestermiller was asking what ΔU is equal to over one complete cycle when you consider ΔU = Q + W. Also, I don't know what you mean by "heat/temperature." They are two different concepts.
 
  • #7
I think @Chestermiller was asking what ΔU is equal to over one complete cycle when you consider ΔU = Q + W. Also, I don't know what you mean by "heat/temperature." They are two different concepts.
Then ΔU is just equal to Q + W? I don't know what to say. And I was asking if it's the difference in either heat or difference in temperature, but I guess it's not. I'm really new to thermodynamics so I'm not sure how to answer these questions. Sorry.

Edit: I tried ot take the difference between high and low temp and use it in the ΔU equation, but I don't know if I even have the right idea yet.
 
  • #8
kuruman
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Then ΔU is just equal to Q + W? I don't know what to say. And I was asking if it's the difference in either heat or difference in temperature, but I guess it's not. I'm really new to thermodynamics so I'm not sure how to answer these questions. Sorry.
That's OK. Can you explain in your own words what each of the symbols ΔU, Q and W stands for? This is not a test. If you are new in thermodynamics, look up the first Law and how to use it in this case.
 
  • #9
That's OK. Can you explain in your own words what each of the symbols ΔU, Q and W stands for? This is not a test. If you are new in thermodynamics, look up the first Law and how to use it in this case.
ΔU is change in energy, in this case thermal energy. Q is heat added to the system. W is the work done.
 
  • #10
kuruman
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ΔU is change in energy, in this case thermal energy. Q is heat added to the system. W is the work done.
More precisely ΔU is the change in internal energy when the system goes from state A to state B, Q is the heat added to the system while the system goes from state A to state B and W is the work done on the system as it goes from state A to state B. What can you tell about these three quantities when the system starts at A and then returns to A after following some unspecified path? By this I mean how are they related to the numbers that are given to you by the problem?
 
  • #11
More precisely ΔU is the change in internal energy when the system goes from state A to state B, Q is the heat added to the system while the system goes from state A to state B and W is the work done on the system as it goes from state A to state B. What can you tell about these three quantities when the system starts at A and then returns to A after following some unspecified path? By this I mean how are they related to the numbers that are given to you by the problem?
If the system starts at A and then returns to A that would mean everything goes back to how it started, right? Does that mean the work out would be the work in? I'm unsure.
 
  • #12
kuruman
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If the system starts at A and then returns to A that would mean everything goes back to how it started, right? Does that mean the work out would be the work in? I'm unsure.
Reason it out. Work in is positive, work out is the negative of work in. The same applies to Q. Positive Q means heat goes in and negative Q means heat comes out. If the work out (done by the engine) and the work in (done on the engine) over a complete cycle have the same magnitude, then the net work over a cycle would be zero. The same can be said about the heat. So if "everything goes back to how it started", there will be no net heat going in and there will be no work done by the heat engine. That's equivalent to having an engine that is not running. Not very useful and certainly not something you want to invest money in. However there is a quantity that returns back to what it was at the beginning of the cycle. What is it?
 
  • #13
Reason it out. Work in is positive, work out is the negative of work in. The same applies to Q. Positive Q means heat goes in and negative Q means heat comes out. If the work out (done by the engine) and the work in (done on the engine) over a complete cycle have the same magnitude, then the net work over a cycle would be zero. The same can be said about the heat. So if "everything goes back to how it started", there will be no net heat going in and there will be no work done by the heat engine. That's equivalent to having an engine that is not running. Not very useful and certainly not something you want to invest money in. However there is a quantity that returns back to what it was at the beginning of the cycle. What is it?
Is the heat going in not the same every cycle?
 
  • #14
kuruman
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Is the heat going in not the same every cycle?
It is. Every cycle is the same as every other cycle.
 
  • #15
It is. Every cycle is the same as every other cycle.
Okay, so how would I find the heat in? Where do I go from here?
 
  • #16
kuruman
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Okay, so how would I find the heat in? Where do I go from here?
You answer the last question in #12. Hint: What is the change in internal energy ΔU over a complete cycle? It might help if you found an expression for it and see what it depends on.
 
  • #17
You answer the last question in #12. Hint: What is the change in internal energy ΔU over a complete cycle? It might help if you found an expression for it and see what it depends on.
Well it has to be either the change in heat or the change in temperature. Or is it a mixture of both? I'm trying to find an expression but most of them deal with the mass of the object.
 
  • #18
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Well it has to be either the change in heat or the change in temperature. Or is it a mixture of both? I'm trying to find an expression but most of them deal with the mass of the object.
Have you studied the Carnot Cycle yet? If so, what is the change in internal energy of the gas when it completes one full cycle? What does that tell you about the net heat added and the net work done?
 

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