Question

1.45 L of 16°C water is placed in a refrigerator. The refrigerator's motor must supply an extra 10.7 W power to chill the water to 6°C in 0.7 hr. What is the refrigerator's coefficient of performance?

Answer 1

The coefficient of performance of the refrigerator is 2.251.

Step-by-step explanation:

In this case, the coefficient of performance of the refrigerator (
`COP`), no unit, is equal to the ratio of the heat rate received from the water to the power needed to work, that is:

`COP = (\dot Q_(L))/(\dot W)` (1)

`COP = (\rho\cdot V\cdot c_(w)\cdot \Delta T)/(\dot W \cdot \Delta t)` (2)

Where:

`\dot Q_(L)` - Heat rate received from the water, in watts.

`\dot W` - Power, in watts.

`\rho` - Density of water, in kilograms per cubic meter.

`V` - Volume of water, in cubic meters.

`c_(w)` - Specific heat of water, in joules per kilogram-degree Celsius.

`\Delta T` - Temperature change, in degrees Celsius.

`\Delta t` - Cooling time, in seconds.

If we know that
`\rho = 1000\,(kg)/(m^(3))`,
`V = 1.45* 10^(-3)\,m^(3)`,
`c_(w) = 4187\,(J)/(kg\cdot ^( \circ)C)`,
`\Delta T = 10\,^(\circ)C`,
`\dot W = 10.7\,W` and
`\Delta t = 2520\,s`, then the coefficient of refrigeration of the refrigerator is:

`COP = (\rho\cdot V\cdot c_(w)\cdot \Delta T)/(\dot W \cdot \Delta t)`

`COP = 2.251`

The coefficient of performance of the refrigerator is 2.251.