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

Answer:

The coefficient of performance of the refrigerator is 2.251.

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:

$COP = \frac{\dot Q_{L}}{\dot W}$ (1)

$COP = \frac{\rho\cdot V\cdot c_{w}\cdot \Delta T}{\dot W \cdot \Delta t}$ (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 $\rho = 1000\,\frac{kg}{m^{3}}$, $V = 1.45\times 10^{-3}\,m^{3}$, $c_{w} = 4187\,\frac{J}{kg\cdot ^{ \circ}C}$, $\Delta T = 10\,^{\circ}C$, $\dot W = 10.7\,W$ and $\Delta t = 2520\,s$, then the coefficient of refrigeration of the refrigerator is:

$COP = \frac{\rho\cdot V\cdot c_{w}\cdot \Delta T}{\dot W \cdot \Delta t}$

$COP = 2.251$

The coefficient of performance of the refrigerator is 2.251.