A Carnot engine operating between 27°C and 127°C has efficienc...
T
1 = (27 + 273)K = 300K
T
2 = (127 + 273)K = 400K
Efficiency,
The correct answer is: 25%
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A Carnot engine operating between 27°C and 127°C has efficienc...
The Carnot engine is a theoretical engine that operates on the Carnot cycle, which consists of two reversible isothermal processes and two reversible adiabatic processes. The efficiency of the Carnot engine is given by:
Efficiency = 1 - (Tc/Th)
where Tc is the temperature of the cold reservoir and Th is the temperature of the hot reservoir.
In this case, the Carnot engine is operating between 27°C and Th.
To calculate the efficiency, we need to convert the temperatures to Kelvin:
Tc = 27 + 273 = 300 K
Let's assume Th is a variable temperature. The efficiency of the Carnot engine will be maximum when Th is at its highest value.
Therefore, to find the maximum efficiency, we need to find the maximum value of Th.
Since the Carnot engine operates between Tc and Th, we can write:
Th = Tc + ΔT
where ΔT is the temperature difference between the hot and cold reservoirs.
The efficiency equation becomes:
Efficiency = 1 - (Tc/Th) = 1 - (Tc/(Tc + ΔT))
To maximize the efficiency, we need to maximize Th, which means maximizing ΔT.
The maximum value of ΔT occurs when Th is at its maximum value, which is the highest temperature achievable.
Therefore, the efficiency of the Carnot engine is determined by the temperature difference ΔT, and the maximum efficiency can only be achieved when the hot reservoir is at its highest possible temperature.