Understanding Heat Engines

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Physics › Understanding Heat Engines

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A Carnot Cycle operates using a heat reservoir at a temperature of and a cold reservoir at a temperature of . What is the operating efficiency, $\eta$ , of this cycle, to the nearest percent?

Explanation

The efficiency of a Carnot Cycle is defined by:

$\eta = 1- \frac{T_C}{T_H}$

In this calculation, temperatures are given in Kelvin. Since Celsius and Kelvin exist in a one-to-one ratio, a conversion is unnecessary unless desired. We will perform the conversion, since most classes will advise it.

$200^oC+273=473K,\ 50^oC+273=323K$

Using these values, we can calculate the efficiency:

$\eta=1-\frac{T_C}{T_H}=1-\frac{323K}{473K}=1-(0.683)=0.317$

Convert this value to a percentage:

$0.317\times 100%=31.7%\approx32%$