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Thermodynamics - Carnot Engine

  • #1

Homework Statement


(a.) For an Ideal Gas Carnot engine, what is the relationship between heat absorbed and rejected by the reservoirs and their temperature? If a heat pump is used to transfer this heat, what is the rate at which heat is added to the hot reservoir in terms of the temperatures of the reservoir and the work done by the pump?

(b.) A house is imperfectly insulated such that it can maintain a heat difference from its surroundings of 5K if it is heated solely by electrical power supplied to electric heaters. If the heaters are turned off and the same power is supplied to a heat pump that extracts heat from a reservoir at the same temperature of the surroundings, 270K, at what temperature can the house be maintained?

Homework Equations


[tex] \frac{Q_H}{Q_C} = \frac{\theta_H}{\theta_C} [/tex]

and

[tex] \frac{Q_H}{W} = \frac{\theta_H}{\theta_H - \theta_C} [/tex]


The Attempt at a Solution



For part (a.) I have the first equation I posted above and

[tex] \frac{dQ_H}{dt} = \frac{\theta_H}{\theta_H - \theta_C} \frac{dW}{dt} [/tex]

which I'm not entirely sure is what the question was looking for

In part (b) I've tried quite a few different things and keep going in circles! I think, with a lot of messy substitution I get the temperature to be 275K. As it was before, which seems consistent with conservation of energy, but I'm not sure its actually right?

This looks like quite a basic question, so its quite embarrassing to be having problems with it, if any one has any helpful advice that would be great. Thanks!
 
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Answers and Replies

  • #2
For example, one of the methods I used was:

[tex] \frac{dQ_H}{dt} = \frac{\theta_H}{\theta_H - \theta_C} \frac{dW}{dt} [/tex]

Which, when [tex] \theta_H = 275 K [/tex] and [tex] \theta_C = 270 K [/tex] give [tex] \frac{dQ}{dt} = \frac{1}{55} \frac{dW}{dt} [/tex]

But then I don't really know what to do with that?
 
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  • #3
anyone?
 
  • #4
i need help with this one too!!
 

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