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wariber [46]
4 years ago
12

An air conditioner is a refrigerator with the inside of the house acting as the cold reservoir and the outside atmosphere acting

as the hot reservoir. Assume that an air conditioner consumes 1.25 * 103 W of electrical power and that it can be idealized as a reversible Carnot refrigerator. If the coefficient of performance of this device is 1.75, how much heat can be extracted from the house in a day
Physics
1 answer:
Monica [59]4 years ago
8 0

Answer:

1.89*10^8 J

Explanation:

The coefficient of performance of this device is \frac{Q}{W} where Q .is the useful heat supplied or removed by the considered system and W  is the work required by the considered system.

Step 1

Coeffiecient of perfromance for cooling (COPC) = 1.75 = \frac{Q}{W} .

                                                     Q = 1.75W

Step 2

We convert day into seconds:

1 day = 24 hrs = 86400 seconds

Step 3

Heat that can be extracted from the house in a day is:

Q = 1.75 * 1.25 * 10^3 * 86400 =  189000000 = 1.89*10^8 J

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v_d^2=u^2+2aS\\\\v=\pm\sqrt{u^2+2aS}\\\\v_d=\pm\sqrt{0+2\times -9.8\times -1.25}\\\\v_d=\pm\sqrt{24.5}=\pm4.95\ m/s

Since, the motion is downward, final velocity must be negative. So,

v_d=-4.95\ m/s

Upward motion:

Given:

Displacement of ball (S) = 0.665 m

Initial velocity (v_d) = 4.95 m/s(Upward direction)

Acceleration is due to gravity (g) = -9.8 m/s²

Using equation of motion, we have:

v_{up}^2=v_d^2+2aS\\\\v_{up}=\pm\sqrt{v_d^2+2aS}\\\\v_{up}=\pm\sqrt{24.5+2\times -9.8\times 0.665}\\\\v_{up}=\pm\sqrt{10.966}=\pm3.31\ m/s

Since, the motion is upward, final velocity must be positive. So,

v_{up}=3.31\ m/s

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Impulse = Final momentum - Initial momentum

J=m(v_{up}-v_d)\\\\J=(0.150\ kg)(3.31-(-4.95))\ m/s\\\\J=0.150\ kg\times 8.26\ m/s\\\\J=1.239\ Ns

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