00:01
This question we have a space heater of power p equals to 3 .00 kilowatt that is 3 ,000 watt and it is used to run to put a kitchen into the same amount of heat as a refrigerator of coefficient of performance 3 .00 does when it freezes mass m equals to 1 .5 kg of water from t initial equals to 20 degrees centigrade and t final that is equal to 0.
00:30
Degree centigrade okay so we have to calculate the total heat first of all so total heat rejected during the free freezing of water at 20 degrees centigrade so it will be equals to m c multiplied by delta t plus m multiplied by latent heat of fuser okay so substituting values so we will get that mass m is 1 .5 multiplied by specific heat is 4186 clok jule per kilogram degree centigrade and delta t it will be 20 minus d d d.
01:00
0 and plus m is 1 .50 and latent heat of fusion which is 3 .3 into 10 to the power 5 jule per kg so from here after solving we get q equals to 620 580 jule okay so we have all now the coefficient of performance k it is equals to q c divided by w so now we can calculate this is uh this heat is rejected uh to the the room, this is heat taken out from the cold reservoir, so which is this value.
01:36
So we can substitute this value as qc.
01:39
So from here, w comes out to be qc by w, qc by k, and k is equal to 3.
01:45
So we can write that it is equal to qc by 3.
01:49
So now we can calculate the heat delivered to the room qh, which will be equals to qc plus w and w is equal to this qc by 3.
01:57
So after solving, we will get 4 by 3 qc...