Solution to 2008 Problem 35

The Carnot efficiency is
\begin{align*}e = 1 - \frac{T_{\text{low}}}{T_{\text{high}}}\end{align*}
where T_{\text{low}} and T_{\text{high}} are the low and high temperatures in ^{\circ} \mbox{ K}. So, in this case,
\begin{align*}e = 1 - \frac{\left(273  +7\right) ^{\circ} \mbox{ K}}{\left(273 + 27\right) ^{\circ} \mbox{ K}} = 1- \frac{280...
The efficiency is defined as the ratio of the work
\begin{align*}e = \frac{W}{Q_{\text{in}}}\end{align*}
where W is the work done by the Carnot engine and Q_{\text{in}} is the heat input to the engine. In this problem, the engine is being run in reverse as a heat pump, so the work W is the work done to the engine and the denominator of e is the heat output by the heat pump.
So, the minimum amount of work required to put out 15000 \mbox{ J} of heat is
\begin{align*}\left(15000 \mbox{ J}\right) \cdot \left(\frac{1}{15} \right) = \boxed{1000 \mbox{ J}}\end{align*}
Therefore, answer (B) is correct.

return to the 2008 problem list

return to homepage

Please send questions or comments to where X = physgre.