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By the end of this section, you will be able to:
  • Describe a refrigerator and a heat pump and list their differences
  • Calculate the performance coefficients of simple refrigerators and heat pumps

The cycles we used to describe the engine in the preceding section are all reversible, so each sequence of steps can just as easily be performed in the opposite direction. In this case, the engine is known as a refrigerator or a heat pump, depending on what is the focus: the heat removed from the cold reservoir or the heat dumped to the hot reservoir. Either a refrigerator or a heat pump is an engine running in reverse. For a refrigerator    , the focus is on removing heat from a specific area. For a heat pump    , the focus is on dumping heat to a specific area.

We first consider a refrigerator ( [link] ). The purpose of this engine is to remove heat from the cold reservoir, which is the space inside the refrigerator for an actual household refrigerator or the space inside a building for an air-conditioning unit.

The figure shows schematic of a refrigerator or heat pump with an upward arrow Q subscript c at T subscript c. When this goes through the refrigerator or pump, an arrow W is added from right and the resultant upward arrow is Q subscript h at T subscript h.
A schematic representation of a refrigerator (or a heat pump). The arrow next to work ( W ) indicates work being put into the system.

A refrigerator (or heat pump) absorbs heat Q c from the cold reservoir at Kelvin temperature T c and discards heat Q h to the hot reservoir at Kelvin temperature T h , while work W is done on the engine’s working substance, as shown by the arrow pointing toward the system in the figure. A household refrigerator removes heat from the food within it while exhausting heat to the surrounding air. The required work, for which we pay in our electricity bill, is performed by the motor that moves a coolant through the coils. A schematic sketch of a household refrigerator is given in [link] .

The figure shows schematic diagram and working of a refrigerator.
A schematic diagram of a household refrigerator. A coolant with a boiling temperature below the freezing point of water is sent through the cycle (clockwise in this diagram). The coolant extracts heat from the refrigerator at the evaporator, causing coolant to vaporize. It is then compressed and sent through the condenser, where it exhausts heat to the outside.

The effectiveness or coefficient of performance     K R of a refrigerator is measured by the heat removed from the cold reservoir divided by the work done by the working substance cycle by cycle:

K R = Q c W = Q c Q h Q c .

Note that we have used the condition of energy conservation, W = Q h Q c , in the final step of this expression.

The effectiveness or coefficient of performance K P of a heat pump is measured by the heat dumped to the hot reservoir divided by the work done to the engine on the working substance cycle by cycle:

K P = Q h W = Q h Q h Q c .

Once again, we use the energy conservation condition W = Q h Q c to obtain the final step of this expression.

Summary

  • A refrigerator or a heat pump is a heat engine run in reverse.
  • The focus of a refrigerator is on removing heat from the cold reservoir with a coefficient of performance K R .
  • The focus of a heat pump is on dumping heat to the hot reservoir with a coefficient of performance K P .

Conceptual questions

If the refrigerator door is left open, what happens to the temperature of the kitchen?

The temperature increases since the heat output behind the refrigerator is greater than the cooling from the inside of the refrigerator.

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Is it possible for the efficiency of a reversible engine to be greater than 1.0? Is it possible for the coefficient of performance of a reversible refrigerator to be less than 1.0?

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Problems

A refrigerator has a coefficient of performance of 3.0. (a) If it requires 200 J of work per cycle, how much heat per cycle does it remove the cold reservoir? (b) How much heat per cycle is discarded to the hot reservoir?

a. 600 J; b. 800 J

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During one cycle, a refrigerator removes 500 J from a cold reservoir and rejects 800 J to its hot reservoir. (a) What is its coefficient of performance? (b) How much work per cycle does it require to operate?

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If a refrigerator discards 80 J of heat per cycle and its coefficient of performance is 6.0, what are (a) the quantity off heat it removes per cycle from a cold reservoir and (b) the amount of work per cycle required for its operation?

a. 69 J; b. 11 J

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A refrigerator has a coefficient of performance of 3.0. (a) If it requires 200 J of work per cycle, how much heat per cycle does it remove the cold reservoir? (b) How much heat per cycle is discarded to the hot reservoir?

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Questions & Answers

What mass of steam of 100 degree celcius must be mixed with 150g of ice at 0 degree celcius, in a thermally insulated container, to produce liquid water at 50 degree celcius
Emmanuel Reply
sorry I dont know
Bamidele
thank you
Emmanuel
What is the pressure?
SHREESH
To convert 0°C ice to 0°c water. Q=M*s=150g*334J/g=50100 J.......... Now 0° water to 50° water... Q=M*s*dt=150g*4.186J/g*50= 31395 J....... Which adds upto 81495 J..... This is amount of heat the steam has to carry. 81495= M *s=M*2230J/g..therefore.....M=36.54g of steam
SHREESH
This is at 1 atm
SHREESH
If there is change in pressure u can refer to the steam table ....
SHREESH
instrument for measuring highest temperature of a body is?
brian Reply
Thermometer
Umar
how does beryllium decay occur
Sandy Reply
Photon?
Umar
state the first law of thermodynamics
Kansiime Reply
Its state that "energy can neither be created nor destroyed but can be transformed from one form to another. "
Ayodamola
what about the other laws can anyone here help with it please
Sandy
The second law of thermodynamics states that the entropy of any isolated system always increases. The third law of thermodynamics states that the entropy of a system approaches a constant value as the temperature approaches absolute zero.
sahil
The first law is very simple to understand by its equation. The law states that "total energy in thermodynamic sytem is always constant" i.e d¶=du+dw where d¶=total heat du=internal energy dw=workdone... PLEASE REFER TO THE BOOKS FOR MORE UNDERSTANDING OF THE CONCEPT.
Elia
what is distance.?
Ali Reply
what is physics?
Ali
Physics is a scientific phenomenon that deals with matter and its properties
Ayodamola
physics is the study of nature and science
John
Chater1to7
min
Physics is branch of science which deals with the study of matters in relation with energy.
Elia
What is differential form of Gauss's law?
Rohit Reply
help me out on this question the permittivity of diamond is 1.46*10^-10.( a)what is the dielectric of diamond (b) what its susceptibility
OLUWA Reply
a body is projected vertically upward of 30kmp/h how long will it take to reach a point 0.5km bellow e point of projection
Abu Reply
i have to say. who cares. lol. why know that t all
Jeff
is this just a chat app about the openstax book?
Lord Reply
kya ye b.sc ka hai agar haa to konsa part
MPL Reply
what is charge quantization
Mayowa Reply
it means that the total charge of a body will always be the integral multiples of basic unit charge ( e ) q = ne n : no of electrons or protons e : basic unit charge 1e = 1.602×10^-19
Riya
is the time quantized ? how ?
Mehmet
What do you meanby the statement,"Is the time quantized"
Mayowa
Can you give an explanation.
Mayowa
there are some comment on the time -quantized..
Mehmet
time is integer of the planck time, discrete..
Mehmet
planck time is travel in planck lenght of light..
Mehmet
it's says that charges does not occur in continuous form rather they are integral multiple of the elementary charge of an electron.
Tamoghna
it is just like bohr's theory. Which was angular momentum of electron is intral multiple of h/2π
Aditya
determine absolute zero
OFERE Reply
The properties of a system during a reversible constant pressure non-flow process at P= 1.6bar, changes from constant volume of 0.3m³/kg at 20°C to a volume of 0.55m³/kg at 260°C. its constant pressure process is 3.205KJ/kg°C Determine: 1. Heat added, Work done, Change in Internal Energy and Change in Enthalpy
Opeyemi Reply
U can easily calculate work done by 2.303log(v2/v1)
Abhishek
Amount of heat added through q=ncv^delta t
Abhishek
Change in internal energy through q=Q-w
Abhishek
please how do dey get 5/9 in the conversion of Celsius and Fahrenheit
Gwam Reply
what is copper loss
timileyin Reply
this is the energy dissipated(usually in the form of heat energy) in conductors such as wires and coils due to the flow of current against the resistance of the material used in winding the coil.
Henry
Practice Key Terms 3

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Source:  OpenStax, University physics volume 2. OpenStax CNX. Oct 06, 2016 Download for free at http://cnx.org/content/col12074/1.3
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