An electroscope is a device that has small gold foil strips suspended from a metal rod. A student brings a negatively charged ro
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Answer:
Thermal Efficiency, η = . . . . . . . . . . . . . . . . Eqn 1
where W₀ = Work Output = Q₁ - Q₀ =82500KW . . . . . . . . . . . . . . . Eqn 2
Q₁ = Heat Supplied/Input = mC(ΔT₁)
Q₁ = Heat Rejected/Output = mC(ΔT₀) . . . . . . . . . . . . . . . . . . . . . . . . Eqn 3
Note: From Carnot's theorem, for any engine working between these two temperatures (T₀/T₁), The maximum attainable efficiency is the Carnot efficiency given as follows;
Therefore, η = 1 - = 1 -
Remember, T₁ = 475K and T₀ = 308K
η = 1 - (308/475) = 1 - 0.648 = 0.352
Hence, the maximum efficiency at which this plant can operate = 35%
2. To determine the minimum amount of rejected heat that must be removed from the condenser every twenty-four hours.
Remember from Eqn 1, Q₁ = W/η,
Therefore, Q₁= 82500/0.35
Q₁=235,714KW,
So, from Eqn 2, Q₀ = 235714 - 82500
Q₀ = 153214KW (KJ/s) (Released Heat)
In t =24 hours, we can then use this to determine the minimum amount of heat rejected qₓ (KiloJoule), = Q₀t (Remember, you have to convert the time, t, unit to seconds)
= 153214 x t (KiloJoule)
qₓ = 153214 x 24 x 3600 (KiloJoule)
qₓ = 13238 MegaJoule
<h3>Therefore, the minimum amount of rejected heat that must be removed from the condenser every twenty-four hours is 13238 MJoule</h3>