A refrigerator has a coefficient of performance cop of 4 it...

A refrigerator has a coefficient of performance (COP) of 4. It can transform 500 g of 25°C water to -10°C ice in 40 minutes. Given the specific heat of water and ice are 4190 J/kg·K and 2110 J/kg·K respectively. Given also the heat of fusion and heat of vaporisation of water are 333 kJ/kg and 2260 kJ/kg respectively. (a) Calculate the heat removed by the refrigerator in the transformation. (b) Calculate the average time rate of electrical energy consumed by the refrigerator in the transformation.

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00:01 In this question, we have given a refrigerator of efficiency 4 and it transformed 500 gram of water into ice from temperature 25 degrees centigrade to temperature minus 10 degrees centigrade.

00:24 And total time it takes during the transformation is 40 minutes.

00:34 We have given the specific heat of water is 4190 jule per kz kelvin.

00:45 And specific heat of ice is also given it is equal to 21110 per kasy kelvin.

00:56 And the latent heat of fusion of water is also given 333 into 10 to the power 3 jule per kizel kelvin.

01:07 So from here, we have to calculate the total energy transformation.

01:16 So, heat evolve during 25 degrees centigrade to 0 degree centigrade.

01:32 It will be equal to mc delta t.

01:37 Mass is given 500 gram when converted into kj20.

01:43 This specific heat of water is also given.

01:49 Change in temperature will be 25 minus 0.

01:56 So from here, q1 is equal to 52 ,375 zoon.

02:05 Now heat evolve during change of state.

02:26 It will be simply equal to m into l.

02:32 Then q2 will be equal to 500 into 10 to the power minus 3 into 33 into 10 to the power 3 it will be equal to 1 ,6665002 now heat evolve during 0 degrees centigrade to minus 10 degree then here mc delta t 500 into 10 to the power minus 3...

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