Thermodynamics - Carnot Engine

In summary, the first part of the conversation discusses the relationship between heat absorbed and rejected by reservoirs and their temperatures in an Ideal Gas Carnot engine. It also mentions the rate at which heat is added to the hot reservoir when using a heat pump. The second part of the conversation discusses a house that is imperfectly insulated and can maintain a heat difference of 5K when heated solely by electric power. It then asks at what temperature the house can be maintained if the same power is supplied to a heat pump extracting heat from a reservoir at 270K. The equations \frac{Q_H}{Q_C} = \frac{\theta_H}{\theta_C} and \frac{Q_H}{W} = \frac{\theta
  • #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 anyone has any helpful advice that would be great. Thanks!
 
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  • #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!
 
  • #5


I can provide a response to the content in the following manner:

For part (a), the relationship between heat absorbed and rejected by the reservoirs and their temperatures is described by the first law of thermodynamics, which states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system. In the case of a Carnot engine, it is an idealized reversible engine that operates between two reservoirs at different temperatures. The heat absorbed by the engine from the hot reservoir is converted into work, while the heat rejected by the engine to the cold reservoir is equal to the work done on the engine. The ratio of these two heat exchanges is given by the Carnot efficiency, which is defined as the ratio of the temperatures of the two reservoirs. In other words, the heat added to the hot reservoir is proportional to its temperature, and the heat rejected by the cold reservoir is proportional to its temperature.

If a heat pump is used to transfer heat from the cold reservoir to the hot reservoir, the rate at which heat is added to the hot reservoir can be calculated using the second law of thermodynamics, which states that the efficiency of a heat pump is less than the Carnot efficiency. This means that for a given amount of work done by the pump, the amount of heat transferred from the cold reservoir to the hot reservoir will be less than the amount of heat rejected by the Carnot engine. The rate at which heat is added to the hot reservoir can be calculated using the formula given in the question, which takes into account the temperature of the hot reservoir and the work done by the pump.

For part (b), the temperature at which the house can be maintained can be calculated using the same principles as in part (a). Since the house is imperfectly insulated, it will lose heat to its surroundings, and this heat loss can be balanced by supplying heat from a heat pump. The temperature at which the house can be maintained will depend on the efficiency of the heat pump and the amount of heat supplied by the pump. Using the equations given in the question, the temperature at which the house can be maintained can be calculated to be 275K, which is consistent with the conservation of energy.
 

1. What is a Carnot engine?

A Carnot engine is a theoretical engine that operates on the principles of thermodynamics. It consists of four main components: a hot reservoir, a cold reservoir, a working substance (usually a gas), and a piston. It is often used as a model for understanding the maximum efficiency of real-world heat engines.

2. How does a Carnot engine work?

A Carnot engine works by taking in heat energy from a hot reservoir and using it to do work. The working substance expands and does work on the piston, which in turn, drives a turbine or other mechanical device. The remaining energy is then expelled as waste heat into the cold reservoir. This process is repeated in a cyclical manner.

3. What is the efficiency of a Carnot engine?

The efficiency of a Carnot engine is given by the Carnot efficiency formula: efficiency = (Thot - Tcold) / Thot, where Thot is the temperature of the hot reservoir and Tcold is the temperature of the cold reservoir. This formula shows that the efficiency of a Carnot engine is dependent on the temperature difference between the two reservoirs.

4. What is the significance of the Carnot cycle?

The Carnot cycle is a theoretical thermodynamic cycle that serves as a benchmark for the maximum efficiency of any heat engine. It shows that the maximum efficiency of a heat engine is dependent only on the temperature difference between the hot and cold reservoirs, and not on the specific working substance or design of the engine.

5. Can a Carnot engine be used in real-world applications?

While a Carnot engine is often used as a theoretical model, it is not commonly used in real-world applications due to practical limitations. The ideal conditions required for a Carnot engine, such as perfectly insulated reservoirs and reversible processes, are difficult to achieve in practice. However, the principles of the Carnot cycle and efficiency are still important concepts in the field of thermodynamics and are used in the design and optimization of real-world heat engines.

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