00:01
Cop of the refrigerator will be heat input, that is actual heat input divided by the power input to the refrigerator.
00:16
So here in the question, heat removing rate is given as 400 kilojoule per minute.
00:22
In watt, it will be 400 kilojoule divided by 60, that is minute.
00:27
So 400 kilojoule or sorry, 400 divided by 60 into kilowatt.
00:35
Whole divided by 3 kilowatt, that is the work consumed or the power input.
00:40
Upon calculation we will have that equals to 2 .22.
00:47
Now there are also the temperature range is given in the question.
00:52
So car not cop or reversible cop will be equals to the ratio of the temperature at which heat is removed.
01:08
That is let's suppose this is the refrigerator and this is the heat input temperature and this is the heat output temperature so it will be cop will be equals to t1 divided by t2 minus t1 so that comes sorry it will be oh no it's okay so minus 5 degrees celsius can be as in kelvin 268 kelvin divided by t2 minus t1 that comes equal to 27.
01:51
Upon calculation we got carnot cop equals to or the reversible cop equals to 9 .925.
02:01
So this is the actual cop 2 .22 and this is the reversible cop 9 .925.
02:11
Now rate of the heat rejection, that is b...