Solving Heat Engine & Refrigerator Problems

In summary, a Carnot heat engine operates between temperatures T1=400K and T2=300K, receiving Qin=1200 J from the high temperature reservoir in each cycle. The heat Qout delivered to the low temperature reservoir can be calculated. When operated in reverse as a refrigerator, the engine receives Qin=1200 J from the low temperature reservoir and the heat delivered to the high temperature reservoir can be determined. The work done by the engine and on the refrigerator can also be calculated, with both answers being positive. The efficiency of the engine and the performance coefficient of the refrigerator can be found using relevant equations.
  • #1
Ki-nana18
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0

Homework Statement


A Carnot heat engine operates between temperatures T1= 400K and T2= 300K. In each cycle the engine receives Qin=1200 J from the high temperature reservoir.
a) Calculate the heat Qout delivered to the low temperature reservoir.
b) Suppose the engine is operated in reverse as a refrigerator. Now the engine receives Qin= 1200 J from the low temperature reservoir. Determine the heat delivered to the high temperature reservoir.
c) Calculate the work done by the engine in part a, and the work done on the refrigerator in part b. Note that both answers should be positive.
d) Find the efficiency of the engine in part a, and the performance coefficient of the refrigerator in part b.

Homework Equations


QH+QC+W=0
##\frac{Q_H} {T_H}##+##\frac{Q_C} {T_C}##=0
e=W/QH
performance coefficient=##\frac{|Q_c|} {W}##

The Attempt at a Solution


[/B]
My confusion arises when I get negative values for part c. Mathematically they should be negative, I suspect that I am not grasping a concept somewhere.
20141004_152957.jpg
 
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  • #2
Let's take the case for the engine first. Look at the diagram you drew at the top. Heat flows from the hot reservoir and a part of it is converted into work done by the engine, while the rest is dumped into the cold reservoir. This statement, translated into mathematical language says ##Q_h=W+Q_c## where ##Q_h,~Q_c,~W## are defined analogously from the above sentence with positive values corresponding to the bolded words. Does this equation make sense to you in terms of the language that it is translated to? Does the sign convention make sense to you?

Now look at the equation you provided in your "relevant equations". You have the quite different equation ##Q_h+Q_c+W=0##, so again, looking at the picture you drew, what English sentence do you think this equation gets translated to?
 
  • #3
Your value for COP, .3333, is not correct. Since it is a Carnot cycle, COP = |Qc/W| = |Qc/(Qh-Qc)| = |Tc/(Th-Tc)|

AM
 

1. How does a heat engine work?

A heat engine works by converting heat energy into mechanical energy. This process involves the transfer of heat from a high temperature reservoir to a low temperature reservoir, and the use of this energy to perform work.

2. What is the Carnot cycle?

The Carnot cycle is a theoretical cycle that describes the most efficient way to convert heat energy into mechanical energy. It consists of four processes: isothermal compression, adiabatic expansion, isothermal expansion, and adiabatic compression.

3. How do you calculate the efficiency of a heat engine?

The efficiency of a heat engine is calculated by dividing the work output by the heat input. This can be expressed as a percentage, with the ideal efficiency for a Carnot engine being (Th - Tc)/Th, where Th is the temperature of the high temperature reservoir and Tc is the temperature of the low temperature reservoir.

4. What is the coefficient of performance (COP) for a refrigerator?

The coefficient of performance (COP) for a refrigerator is a measure of its efficiency in removing heat from a cold reservoir and transferring it to a hot reservoir. It is calculated by dividing the heat removed from the cold reservoir by the work required to do so.

5. How can the efficiency of a refrigerator be improved?

The efficiency of a refrigerator can be improved by increasing the temperature difference between the hot and cold reservoirs, using more efficient refrigerants, and reducing energy losses through insulation and other design improvements. Additionally, regular maintenance and proper use can also help to improve efficiency.

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