The drinking water needs of an office are met by cooling tab water in a refrigerated water fountain from 23 to 6°C at an average rate of 10 kg/h. If the COP of this refrigerator is 3.1, the required power input to this refrigerator is

(a) 197W
(b) 612W
(c) 64W
(d) 109W
(e) 403W

The correct option is;

(c) 64W

Explanation:

Here we have the Coefficient Of Performance, COP given by

The heat change from 23° to 6°C for a mass of 10 kg/h which is equivalent to 10/(60×60) kg/s or 2.78 g/s we have

= m·c·ΔT = 2.78 × 4.18 × (23 - 6) = 197.39 J

Therefore, plugging in the value for   in the COP equation we get;

which gives

Since we were working with mass flow rate then the power input is the same as the work done per second and the power input to the refrigerator = 63.674 J/s ≈ 64 W.

The power input to the refrigerator is approximately 64 W.

b

step-by-step explanation: a third of 24 is 8, 8 times 3 is 24 so 8 - 2 would be 6

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