The refrigerator's coefficient of performance is 6.
The heat extracted from the cold reservoir Q cold (i.e., inside a refrigerator) divided by the work W required to remove the heat is known as the coefficient of performance, or COP, of a refrigerator (i.e., the work done by the compressor). The required inside temperature and the outside temperature have a significant impact on the COP.
As the inside temperature of the refrigerator decreases, its coefficient of performance decreases. The coefficient of performance (COP) of refrigeration is always more than 1.
The heat produced in the cold compartment, H = 780.0 J
Work done in ideal refrigerator, W = 130.0 J
Refrigerator's coefficient of performance = H/W
= 780/130
= 6
Therefore, the refrigerator's coefficient of performance is 6.
Energy conservation requires the exhaust heat to be = 780 + 130
= 910 J
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There are missing data in the text of the problem (found them on internet):
- speed of the car at the top of the hill:

- radius of the hill:

Solution:
(a) The car is moving by circular motion. There are two forces acting on the car: the weight of the car

(downwards) and the normal force N exerted by the road (upwards). The resultant of these two forces is equal to the centripetal force,

, so we can write:

(1)
By rearranging the equation and substituting the numbers, we find N:

(b) The problem is exactly identical to step (a), but this time we have to use the mass of the driver instead of the mass of the car. Therefore, we find:

(c) To find the car speed at which the normal force is zero, we can just require N=0 in eq.(1). and the equation becomes:

from which we find