Question

a refrigerator has to transfer an average of 263 j of heat from _ 10 to 25 centigrade calculate the average power consumed assumed that the refrigerator is ideal

Answer 1

WHERE IS SOLUTION OF THIS PROBLEM

Answer 2

263j each second?

Answer 3

This is a basic problem in thermodynamics.
Because we are allowed to regard the refrigerator as ideal, we can treat its operation in the same way as a Carnot engine.
This means that the heat taken from the cold temperature, Q2, and the heat delivered to the higher temperature, Q1, are in the ratio T2 to T1, as long as we express the temperatures in Kelvin.\[\frac{ Q2 }{ Q1}=\frac{ T2 }{ T1 }\]
If we define W to be the work done in order to run the refrigerator to cause the heat transfer Q2, then conservation of energy gives\[Q1=Q2+W\]
With a bit of rearranging you can find\[W=\frac{ Q2 }{ T2 }\left( T1-T2 \right)\]
Assuming the 263 Joules mentioned in the question is the heat transfer per second, you can use the above equation to find the power consumption of the refrigerator.
I made it about 35 Watts.