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

Answer:

700 minutes

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.