An ideal heat engine has a certain efficiency. Two types of changes to the engine operation are considered.

| September 19, 2021 | education

An ideal heat engine has a certain efficiency. Two types of changes to the engine operation are considered. Once the temperature of the heater rises by an amount at a constant temperature of the refrigerator, the second time the temperature of the refrigerator is lowered by an amount, leaving the same temperature of the heater. In which case, the efficiency of the new heat engine will be higher in the first, in the second, in both cases, the same answer depends on the initial temperatures of the heater and refrigerator

The efficiency of a heat engine is determined by the formula: η = 1- To / Tn where To is the temperature of the cooler, Tn is the temperature of the heater. Then the change in efficiency in the first case is equal to:
Δη1 = (1-To / Tn) – (1-To / Tn + ΔT) = To / (Tn + ΔT) -To / Tn
and in the second case:
Δη2 = (1-To / Tn) – (1- (To + ΔT) / Tn)) = (To + ΔT) / Tn-To / Tn
Let’s find the difference between these changes:
Δη1-Δη2 = To / (Tn + ΔT) – (Tn + ΔT) / Tn = (To * Tn- (Tn + ΔT) (Tn + ΔT)) / (Tn + ΔT) * Tn = (To * Tn- (Tn + ΔT) ^ 2) / (Tn + ΔT) * Tn.
The third answer is obvious.



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