Solution to 1992 Problem 16

The well-known formula for the efficiency of an ideal Carnot engine is
\begin{align*}e = 1 - \frac{T_L}{T_H}\end{align*}
where T_L is the temperature of the low-temperature reservoir and T_H is the temperature of the high-temperature reservoir (see page 522 of Giancoli). Both temperatures MUST be in Kelvin. The efficiency is DEFINED as
\begin{align*}e = \frac{\left|W\right|}{\left|Q_H\right|}\end{align*}
where W is the work performed by the engine and Q_H is the heat absorbed from the high-temperature body (see Giancoli page 519). Therefore, if 2000 \mbox{ J} are input into the engine, then the amount of work that the engine performs is
\begin{align*}\left(2000 \mbox{ J}\right) e = \left(2000 \mbox{ J}\right) \left(1 - \frac{527 \mbox{ K}+ 273 \mbox{ K}}{727 \...
Therefore, answer (A) is correct.

return to the 1992 problem list

return to homepage

Please send questions or comments to where X = physgre.