Respuesta :
Answer:
The efficiency of this ideal and reversible engine is 85 percent.
The efficiency of the Carnot cycle represents the efficiency of a thermal machine with no irreversibilities, hence, it is impossible for any real engine operating between the two reservoirs cannot be more efficient than this engine.
Explanation:
Let assume that the temperature of the atmosphere is 300 K. From Thermodynamics we know that the efficiency of the Carnot's cycle ([tex]\eta_{th}[/tex]), dimensionless, is:
[tex]\eta_{th} = 1-\frac{T_{L}}{T_{H}}[/tex] (1)
Where:
[tex]T_{H}[/tex] - Temperature of the kerosene combustor (hot reservoir), measured in kelvins.
[tex]T_{L}[/tex] - Temperature of the atmosphere (cold reservoir), measured in kelvins.
If we know that [tex]T_{L} = 300\,K[/tex] and [tex]T_{H} = 2000\,K[/tex], then the efficieny of this ideal and reversible engine is:
[tex]\eta_{th} = 1-\frac{300\,K}{2000\,K}[/tex]
[tex]\eta_{th} = 0.85[/tex]
The efficiency of this ideal and reversible engine is 85 percent.
The efficiency of the Carnot cycle represents the efficiency of a thermal machine with no irreversibilities, hence, it is impossible for any real engine operating between the two reservoirs cannot be more efficient than this engine.