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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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Calculate the magnitude of the electric field inside the solid at a distance of 9.50 cm from the center of the cavity. Express y
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Question:

A point charge of -2.14uC  is located in the center of a spherical cavity of radius 6.55cm  inside an insulating spherical charged solid. The charge density in the solid is 7.35×10−4 C/m^3.

a) Calculate the magnitude of the electric field inside the solid at a distance of 9.50cm  from the center of the cavity.  

Express your answer using two significant figures.

Answer:

The magnitude of the electric field inside the solid at a distance of 9.50cm from the center of the cavity 3.65\times 10^5N/C

Explanation:

A point charge ,q = -2.14\times 10^{-6} C is located in the center of a spherical cavity of radius , r =6.55\times 10^{-2}  m inside an insulating spherical charged solid.  

The charge density in the solid , d = 7.35 \times 10^{-4}C/m^3.

Distance from the center of the cavity,R =9.5\times 10^{-2 }m

Volume of shell of charge= V  =(\frac{4\pi}{3})[ R^3 - r^3 ]

Charge on the shell ,Q = V \times d'

Q =(\frac{4\pi}{3})[ R^3 - r^3 ] \times d

Q = 4.1888*\times 10^{-4 }[8.57375 - 2.81011 ]\times 7.35\times 10^{-4}

Q = 4.1888\times 10^{-4} [5.76364 ] \times 7.35 \times 10^{-4}

Q =2.4143 \times 10^{-4} \times 7.35 \times 10^ { -4}

Q =1.7745 \times 10^{-6 }C

Electric field at 9.5\times 10^{-2}m due to shellE1  = \frac{k Q}{R^2}

E1 =  \frac{ 9 \times 10^9\times 1.7745\times 10^{-6 }}{ 90.25\times 10^{-4}}

E1 =1.769\times 10^6 N/C

Electric field at  9.5\times 10^{-2} due to 'q' at center E2 = \frac{kq}{R^2}

E2 =\frac{ - 9 \times 10^9\times 2.14\times 10^{-6 }}{ 90.25\times 10^{-4}}

E2 =2.134\times 10^6 N/C

The magnitude of the electric field inside the solid at a distance of 9.50cm from the center of the cavity

= E2- E1

=[  2.134  - 1.769 ]\times 10^6

= 3.65\times 10^5 N/C

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