An engine operates between temperatures of \(650 \mathrm{K}\) and $350 \mathrm{K}\( at \)65.0 \%$ of its maximum possible efficiency. (a) What is the efficiency of this engine? (b) If $6.3 \times 10^{3} \mathrm{J}$ is exhausted to the low temperature reservoir, how much work does the engine do?

The maximum possible efficiency of a heat engine operating between two temperatures can be given by the Carnot efficiency formula: Carnot efficiency = \(1 - \frac{T_{low}}{T_{high}}\), where \(T_{low}\) is the temperature of the low-temperature reservoir, and \(T_{high}\) is the temperature of the high-temperature reservoir. Substituting the given values in the formula, we get: Carnot efficiency = \(1 - \frac{350 K}{650 K} = 1 - 0.5385\).

The engine operates at \(65.0\%\) of its maximum possible efficiency (Carnot efficiency). So, to calculate the actual efficiency of the engine, multiply the percentage by Carnot efficiency: Actual efficiency = \((0.650)(1 - 0.5385)\) Actual efficiency = \(0.29945\) or \(29.945 \%\)

The definition of efficiency for a heat engine is as follows: Efficiency = \(\frac{Work Output}{Heat Input}\) Rearranging for work output, we get: Work Output = Efficiency × Heat Input We are given the heat input to the low-temperature reservoir, which is represented by \(Q_{low}\) = \(6.3 \times 10^{3} J\). Since \(Q_{high} = Q_{low} + Work Output\), we can write the equation for efficiency as: Efficiency = \(\frac{Work Output}{Q_{high}-Q_{low}}\) Rearranging for work output, we get: Work Output = Efficiency × (\(Q_{high} - Q_{low}\)) The difference between heat input and output is based on the engine efficiency. We can rewrite the equation as: Work Output = \(Q_{low} \times \frac{Efficiency}{1 - Efficiency}\) Plugging the values, we get: Work Output = \((6.3 \times 10^{3} J) \times \frac{0.29945}{1 - 0.29945}\) Work Output ≈ \(2.7 \times 10^3 J\) Thus, the engine does approximately \(2.7 \times 10^3 J\) of work.