Physics Help Forum Thermodynamic

 Feb 26th 2018, 12:25 AM #1 Junior Member   Join Date: Feb 2018 Posts: 4 Thermodynamic The heat capacity of the interior of a refrigerator is 4.2 / kJ K . The minimum work that must be done to lower the internal temperature from 18 degree C to 17 degree C, when the outside temperature is 27 degree C will be
 Feb 26th 2018, 12:43 AM #2 Senior Member   Join Date: Apr 2015 Location: Somerset, England Posts: 1,035 What have you done so far? topsquark likes this.
 Feb 26th 2018, 03:01 AM #3 Junior Member   Join Date: Feb 2018 Posts: 4 Hi, I am confused with ambient(Outer Temperature). Specific heat is given So Q=mc(t2-t1) So is Q is min W?
 Feb 26th 2018, 03:58 AM #4 Senior Member   Join Date: Apr 2015 Location: Somerset, England Posts: 1,035 Good, you are interested. Look at the units of heat capacity. I don't think you need the mass of the refrigerator contents, and you are not given it. The heat capacity refers to the entire contents, not per kilogramme. Doing it this way is very common in engineering. Then you will have calculated the heat transferred to lower the temperature, but you are asked for the work done. Where is the heat transferred to? and what is its temperature? So the refrigerator is acting as a heat engine, doing work transferring heat from a colder to hotter body. What do you know about the efficiency of such a heat engine? Do you know any (simple) formulae?
 Feb 26th 2018, 04:29 AM #5 Junior Member   Join Date: Feb 2018 Posts: 4 Efficiency is simply T1-T2/T2 What is the role of Q. And what is the role of outside temperature is 27 degree C
Feb 26th 2018, 05:03 AM   #6
Senior Member

Join Date: Apr 2015
Location: Somerset, England
Posts: 1,035
 Efficiency is simply T1-T2/T2
Yes so what do you think T1 and T2 are?

And do you not know any other details about efficiency?
For instance output/input (perhaps x100 if expressed as a % )

 Feb 26th 2018, 12:58 PM #7 Senior Member   Join Date: Apr 2015 Location: Somerset, England Posts: 1,035 Read this from Wiki Coefficient of performance The efficiency of a refrigerator or heat pump is given by a parameter called the coefficient of performance (COP). The COP of a refrigerator is given by the following equation: COP = Desired Output/Required Input = Cooling Effect/Work Input = QL/Wnet,in The COP of a heat pump is given by the following equation: COP = Desired Output/Required Input = Heating Effect/Work Input = QH/Wnet,in Both the COP of a refrigerator and a heat pump can be greater than one. Combining these two equations results in: COPHP = COPR + 1 for fixed values of QH and QL This implies that COPHP will be greater than one because COPR will be a positive quantity. In a worst-case scenario, the heat pump will supply as much energy as it consumes, making it act as a resistance heater. However, in reality, as in home heating, some of QH is lost to the outside air through piping, insulation, etc., thus making the COPHP drop below unity when the outside air temperature is too low. Therefore, the system used to heat houses uses fuel.[2] For an ideal refrigeration cycle: COP = TL/(TH-TL) For an ideal heat pump cycle: COP = TH/(TH-TL) For Carnot refrigerators and heat pumps, COP is expressed in terms of temperatures: COPR,Carnot = 1/((TH/TL) - 1) COPHP,Carnot = 1/(1 - (TL/TH)) You have two equations for efficiency (called coefficient of performance, COP Equate these to find the work done. topsquark likes this.