Which formula gives the efficiency of a heat engine in terms of hot and cold reservoir temperatures, using Kelvin?

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Multiple Choice

Which formula gives the efficiency of a heat engine in terms of hot and cold reservoir temperatures, using Kelvin?

Explanation:
The maximum possible efficiency of a heat engine operating between two reservoirs depends on how hot the engine’s input is compared to the cold sink, and for an ideal (reversible) case this is given by the Carnot expression: η = 1 − T_cold/T_hot, which is the same as (T_hot − T_cold)/T_hot when temperatures are in Kelvin. Using Kelvin matters because it uses absolute temperatures; with Celsius, the zero point isn’t absolute and the ratio wouldn’t correctly reflect the true thermodynamic limits. So the correct form is the one that directly expresses the fraction of heat converted to work as the difference between the hot and cold temperatures divided by the hot temperature. As a quick check, if T_hot = 500 K and T_cold = 300 K, η = (500 − 300)/500 = 0.4, or 40%. The other forms would give nonsensical results for a real engine (often exceeding 1 or yielding a negative efficiency) and thus don’t represent the thermodynamic limit.

The maximum possible efficiency of a heat engine operating between two reservoirs depends on how hot the engine’s input is compared to the cold sink, and for an ideal (reversible) case this is given by the Carnot expression: η = 1 − T_cold/T_hot, which is the same as (T_hot − T_cold)/T_hot when temperatures are in Kelvin. Using Kelvin matters because it uses absolute temperatures; with Celsius, the zero point isn’t absolute and the ratio wouldn’t correctly reflect the true thermodynamic limits.

So the correct form is the one that directly expresses the fraction of heat converted to work as the difference between the hot and cold temperatures divided by the hot temperature. As a quick check, if T_hot = 500 K and T_cold = 300 K, η = (500 − 300)/500 = 0.4, or 40%.

The other forms would give nonsensical results for a real engine (often exceeding 1 or yielding a negative efficiency) and thus don’t represent the thermodynamic limit.

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