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Vadim26 [7]
2 years ago
10

Electrical energy is used to turn the blades of a fan. The amount of energy transformed is seen here : 750 J electrical energy i

s transformed into 400 J kinectic energy . What happened to the remaining 350j of energy?
Physics
1 answer:
Charra [1.4K]2 years ago
3 0

Electrical energy is used to run the fan

Here as per given condition 750 J of electrical energy is used to run the fan which is converted into Kinetic energy as 400 J

So here we can see that 350 J of energy is lost against many other type of frictional and resistive loses.

So here we can say that out of 750 J of energy only 400 J is used to run the fan and rest amount of energy is lost against friction.

also we can say that efficiency of this fan will be

\eta = \frac{Useful}{Total}

\eta = \frac{400}{750}

\eta = \frac{8}{25}


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3 years ago
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Answer:

a) k_{avg}=6.22\times 10^{-21}

b) k_{avg}=8.61\times 10^{-21}

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Explanation:

Average translation kinetic energy (k_{avg}) is given as

k_{avg}=\frac{3}{2}\times kT    ....................(1)

where,

k = Boltzmann's constant ; 1.38 × 10⁻²³ J/K

T = Temperature in kelvin

a) at T = 27.8° C

or

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substituting the value of temperature in the equation (1)

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k_{avg}=\frac{3}{2}\times 1.38\times 10^{-23}\times 300.8  

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b) at T = 143° C

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T = 143 + 273 = 416 K

substituting the value of temperature in the equation (1)

we have

k_{avg}=\frac{3}{2}\times 1.38\times 10^{-23}\times 416  

k_{avg}=8.61\times 10^{-21}J

c ) The translational kinetic energy per mole of an ideal gas is given as:

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now at T = 27.8° C

        k_{mol}=6.02\times 10^{23}\times 6.22\times 10^{-21}

          k_{mol}=3.74\times 10^{3}J/mol

d) now at T = 143° C

        k_{mol}=6.02\times 10^{23}\times 8.61\times 10^{-21}

          k_{mol}=5.1\times 10^{3}J/mol

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