Question

A refrigerator is to cool 50 kg of water from 250C to 50C. The specific heat of water is 4.2 kJ/kgK. If the COP of the refrigerator is 2.0 and has a power input of 500 W, find the time taken to cool the water.

Answer 1

700 minutes

Step-by-step explanation:

Heat loss by water = Heat gained by the refrigerator

`m_(w)`
`c_(w)`ΔT = IVt

But, power = IV

`m_(w)`
`c_(w)`ΔT = Gt

G = (P x COP)

Thus,

`m_(w)`
`c_(w)`(
`T_(2)` -
`T_(1)`) = (P x COP) x t

where
`m_(w)` is the mass of water,
`c_(w)` is the specific heat of water, ΔT = (
`T_(2)` -
`T_(1)`) is the change in temperature, P is the power of the refrigerator in kilowatts and t is the time taken.

`m_(w)` = 50 kg,
`c_(w)` = 4.2 kJ/(kg K),
`T_(2)` = 250
`^(o)C` (523 K),
`T_(1)` = 50
`^(o)C`, P = 500W (0.5 KW), COP = 2.0 KW

`m_(w)`
`c_(w)`(
`T_(2)` -
`T_(1)`) = (P x COP) x t

50 x 4200 x (523 - 323) = (0.5 x 2.0) x t

50 x 4.2 x 200 = t

42000 = t

⇒ t = 42000

= 42000 s

t = 700 minutes

The time taken to cool the water is 700 minutes.