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Rainbow [258]
3 years ago
7

heat engine operating between 100 °C and 700 °C has an efficiency equal to 40% of the maximum theoretical efficiency. How much e

nergy does this engine extract from the hot reservoir in order to do 5000 J of mechanical work?
Physics
1 answer:
ipn [44]3 years ago
3 0

Answer:

Explanation:

Theoretical efficiency = T₁ - T₂ / T₁ where T₁ and T₂ is absolute temperature of hot and cold end of the heat engine.

= 600 / (273 + 700 )

= 600 / 973

= .6166

operating efficiency = 40% of .6166

= .4 x .6166

= .2466 = 24.66 %

efficiency = work output / heat input

= 5000 / heat input = .2466

heat input = 5000 / .2466

= 20275.75 J .

HEAT EXTRACED = 20275.75 J.

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Answer:

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Explanation:

from the question we are told that

  The displacement of a mass on a spring in simple harmonic motion is A/2 from the equilibrium position

Generally the total mechanical  energy of the mass is mathematically represented as

        TM  =  \frac{1}{2}  *  k  *  A^2

Here  k is the spring constant  ,  A is the total displacement of the  the mass  from maximum  compression to maximum extension of the spring

Generally this total mechanical energy is mathematically represented as

        TM  =  KE  + PE

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Here the potential  energy of the mass is mathematically represented as

     PE   = \frac{1}{ 2}  *  k *  [ x ]^2

Here x is the displacement of the mass from maximum compression or extension of the spring to equilibrium position and the value is  

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So

     PE   = \frac{1}{ 2}  *  k *  [ \frac{A}{2}  ]^2

So

      KE =  \frac{1}{2}  *  k  *  A^2 - \frac{1}{2}  *  k  *  [\frac{A}{2} ]^2

=>    KE =  \frac{1}{2}  *  k  *  A^2 - \frac{1}{8}  *  k  *  A ^2

=>    KE =  0.375  *  k  *  A^2

So the ratio of  KE :  TM is  mathematically represented as

       \frac{KE}{TM} =  \frac{0.375  k A^2 }{0.5 k A^2}

=>    \frac{KE}{TM} = 0.75

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