Answer :
An ideal Carnot engine has an efficiency of 83.0% and performs 4500 J of work every cycle. The energy that is discharged to the lower temperature reservoir every cycles 920J.
The efficiency of an ideal Carnot engine is given by:
efficiency = 1 - (T_cold/T_hot) where T_cold is the temperature of the lower temperature reservoir and T_hot is the temperature of the higher temperature reservoir. From the given efficiency of 83%, we can write:
0.83 = 1 - (T_cold/T_hot). Rearranging this equation, we get:
T_cold/T_hot = 0.17. The ratio of the temperatures is 0.17.Let the energy discharged to the lower temperature reservoir every cycle be Q_cold. The work done by the engine every cycle is 4500 J.
According to the first law of thermodynamics:
Q_hot - Q_cold = 4500 J where Q_hot is the energy absorbed from the higher temperature reservoir every cycle. Using the equation for the ratio of temperatures, we can write:
Q_cold/Q_hot = 0.17.
Rearranging this equation, we get:
Q_cold = 0.17 Q_hot
Substituting this into the first law equation, we get:
Q_hot - 0.17Q_hot = 4500 J.
Simplifying this equation, we get:
0.83Q_hot = 4500 JQ_hot = 4500 J/0.83 = 5421.69 J.
Therefore, the energy discharged to the lower temperature reservoir every cycle is:Q_cold = 0.17Q_hot = 0.17(5421.69 J) ≈ 920 J.
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