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MrRa [10]
4 years ago
14

The hot reservoir for a Carnot engine has a temperature of 889 K, while the cold reservoir has a temperature of 657 K. The heat

input for this engine is 4710 J. The 657-K reservoir also serves as the hot reservoir for a second Carnot engine. This second engine uses the rejected heat of the first engine as input and extracts additional work from it. The rejected heat from the second engine goes into a reservoir that has a temperature of 406 K. Find the total work delivered by the two engines.
Physics
1 answer:
SVETLANKA909090 [29]4 years ago
7 0

Answer:

Explanation:

Efficiency of first engine

= T₁ - T₂ / T₁  where T₁ is temperature of hot reservoir , T₂ is temperature of cold reservoir

= (889 - 657 ) / 889

= ,261 or 26.1 %

output of work = .261 x 4710 = 1229.15 J .

Efficiency of second engine

=  (657 - 406 ) / 657

= .382

Heat rejected in engine one is heat input of second engine

heat input of second engine = 4710 - 1229.15 = 3480.85

output of work of second engine

=  .382 x 3480.85 = 1329.68 J

Total work delivered by two engine

= 1229.15  +  1329.68 J

= 2558.83 J

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3 years ago
The magnetic flux through a coil of wire containing two loops changes at a constant rate from -58 wb to 38 wb in 0. 34 s. What i
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The emf induced in the coil is -5.65 V

<h3>Induced emf in coil</h3>

The induced emf in the coil is given by ε = -NΔΦ/Δt where

  • ΔΦ = change in magnetic flux Φ₂ - Φ₁ where
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  • Φ₂ = final magnetic flux = 38 Wb and and
  • Δt = change in time = t₂ - t₁ where
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Since ε = -NΔΦ/Δt

ε = -N(Φ₂ - Φ₁)/(t₂ - t₁)

Substituting the values of the variables into the equation, we have

ε = -N(Φ₂ - Φ₁)/(t₂ - t₁)

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ε = -5.65 Wb/s

ε = -5.65 V

So, the emf induced in the coil is -5.65 V

Learn more about induced emf in coil here:

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