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A Carnot engine that uses a high temperature reservoir of 800 K has an efficiency of 40%. In order for the efficiency to increase to 50%, the temperature of the high temperature reservoir is increased to?

Sagot :

MathPhys

Answer:

960 K

Explanation:

The efficiency (η) of a Carnot engine is equal to one minus the ratio of the cold temperature (Tc) to the hot temperature (Th).

[tex]\Large \text{$ \eta=1- $}\huge \text{$ \frac{T_C}{T_H} $}[/tex]

Initially, the high temperature is 800 K and the efficiency is 40%. Therefore, the cold temperature is:

[tex]\Large \text{$ 0.40=1- $}\huge \text{$ \frac{T_C}{800\ K} $}\\\\\Large \text{$ 0.60= $}\huge \text{$ \frac{T_C}{800\ K} $}\\\\\Large \text{$ T_C=480\ K $}[/tex]

To increase the efficiency to 50%, the high temperature must become:

[tex]\Large \text{$ 0.50=1- $}\huge \text{$ \frac{480\ K}{T_H} $}\\\\\Large \text{$ 0.50=\ $}\huge \text{$ \frac{480\ K}{T_H} $}\\\\\Large \text{$ T_H=960\ K $}[/tex]

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