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barxatty [35]
3 years ago
11

A 20 watt light bulb is left burning inside a refrigerator operating on a reverse Carnot cycle. If the refrigerator also draws 2

0 watts of electrical power to operate its condenser, can it cool its interior below room temperature while the bulb is on? Explain why or why not
Physics
1 answer:
Gwar [14]3 years ago
8 0

Explanation:

Given that,

Power drawn by the refrigerator = 20 watt

Power of burning light inside = 20 watt

Room temperature = 27°C

We need to calculate the temperature

Using formula of coefficient of refrigerator

COR_{R}=\dfrac{Q_{c}}{W}

\dfrac{Q_{c}}{W}=\dfrac{T_{c}}{T_{H}-T_{c}}

Put the value into the formula

\dfrac{20}{20}=\dfrac{T_{c}}{T_{H}-T_{c}}

\dfrac{T_{c}}{T_{H}-T_{c}}=1

T_{h}=2T_{c}

T_{c}=\dfrac{T_{h}}{2}

T_{c}=\dfrac{27+273}{2}

T_{c}=150\ K

T_{c} is less than T_{h}

Therefore, refrigerator cools its interior below room temperature while the bulb is on.

Hence, This is the required solution.

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Answer:

R=m*g-∀fl*g*l3

Explanation:

<em>An iron block of density rhoFe and of volume l 3 is immersed in a fluid of density rhofluid. The block hangs from a scale which reads W as the weight. The top of the block is a height h below the surface of the fluid. The correct equation for the reading of the scale is</em>

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