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?

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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?

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Answer:The coefficient of performance of the refrigerator is 2.251.Explanation:In this case, the coefficient of performance of the refrigerator ([tex]COP[/tex]), no unit, is equal to the ratio of the heat rate received from the water to the power needed to work, that is:[tex]COP = \frac{\dot Q_{L}}{\dot W}[/tex] (1)[tex]COP = \frac{\rho\cdot V\cdot c_{w}\cdot \Delta T}{\dot W \cdot \Delta t}[/tex] (2)Where:[tex]\dot Q_{L}[/tex] - Heat rate received from the water, in watts.[tex]\dot W[/tex] - Power, in watts. [tex]\rho[/tex] - Density of water, in kilograms per cubic meter.[tex]V[/tex] - Volume of water, in cubic meters.[tex]c_{w}[/tex] - Specific heat of water, in joules per kilogram-degree Celsius.[tex]\Delta T[/tex] - Temperature change, in degrees Celsius.[tex]\Delta t[/tex] - Cooling time, in seconds.If we know that [tex]\rho = 1000\,\frac{kg}{m^{3}}[/tex], [tex]V = 1.45\times 10^{-3}\,m^{3}[/tex], [tex]c_{w} = 4187\,\frac{J}{kg\cdot ^{ \circ}C}[/tex], [tex]\Delta T = 10\,^{\circ}C[/tex], [tex]\dot W = 10.7\,W[/tex] and [tex]\Delta t = 2520\,s[/tex], then the coefficient of refrigeration of the refrigerator is:[tex]COP = \frac{\rho\cdot V\cdot c_{w}\cdot \Delta T}{\dot W \cdot \Delta t}[/tex][tex]COP = 2.251[/tex]The coefficient of performance of the refrigerator is 2.251.

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