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Eduardwww [97]
2 years ago
8

A typical electric refrigerator has a power rating of 500 Watts, which is the rate (J/s) at which electrical energy is supplied

to do work needed to remove heat from the refrigerator. If the refrigerator releases heat to the room at a rate of 800 Watts, at what rate (in Watts) does it remove heat from inside of the refrigerator?
Physics
1 answer:
Goshia [24]2 years ago
3 0

Answer:

The rate of heat removed from inside the refrigerator is 300 watts.

Explanation:

By the First Law of Thermodynamics and the definition of a Refrigeration Cycle, we have the following formula to determine the rate of heat removed from inside the refrigerator (\dot Q_{L}), in watts:

\dot Q_{L} = \dot Q_{H}-\dot W (1)

Where:

\dot Q_{H} - Rate of heat released to the room, in watts.

\dot W - Rate of electric energy needed by the refrigerator, in watts.

If we know that \dot Q_{H} = 800\,W and \dot W = 500\,W, then the rate of heat removed from inside the refrigerator is:

\dot Q_{L} = \dot Q_{H}-\dot W

\dot Q_{L} = 300\,W

The rate of heat removed from inside the refrigerator is 300 watts.

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<em>work done in pumping the entire fuel is 466587 J</em>

<em></em>

Explanation:

weight of the gasoline per volume = 6600 N/m^3

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length of the tank = 2 m

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work done in pumping fuel to this height = ?

First, we find the volume of the fuel

since the tank is cylindrical,<em> we assume that the fuel within also takes the cylindrical shape.</em>

<em>Also, we assume that the fuel completely fills the tank.</em>

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work done = (total weight of the fuel) x (height through which the fuel is pumped)

work done in pumping = 93317.4 x 5 = <em>466587 J</em>

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