A heat engine takes in \(125 \mathrm{kJ}\) of heat from a reservoir at $815 \mathrm{K}\( and exhausts \)82 \mathrm{kJ}\( to a reservoir at \)293 \mathrm{K}$ (a) What is the efficiency of the engine? (b) What is the efficiency of an ideal engine operating between the same two reservoirs?

To find the work output of the given heat engine, subtract the heat exhaust from the heat input: \(W = Q_{input} - Q_{exhaust}\) \(W = 125\,\text{kJ} - 82\,\text{kJ}\) \(W = 43\,\text{kJ}\)

To find the efficiency of the given heat engine, divide the useful work output (Step 1) by the heat input: \(\text{Efficiency} = \frac{W}{Q_{input}}\) \(\text{Efficiency} = \frac{43\,\text{kJ}}{125\,\text{kJ}}\) \(\text{Efficiency} = 0.344\) To express this as a percentage, multiply by 100: \(\text{Efficiency} = 34.4\%\)

The efficiency of an ideal (Carnot) engine is calculated using the temperatures of the two reservoirs: \(\text{Carnot Efficiency} = 1 - \frac{T_{cold}}{T_{hot}}\) where \(T_{hot}\) is the temperature of the hot reservoir and \(T_{cold}\) is the temperature of the cold reservoir (both in Kelvins). \(\text{Carnot Efficiency} = 1 - \frac{293\,\text{K}}{815\,\text{K}}\) \(\text{Carnot Efficiency} = 0.6402\) To express this as a percentage, multiply by 100: \(\text{Carnot Efficiency} = 64.02\%\) Now we have the solutions to both parts (a) and (b) of the exercise. (a) The efficiency of the given heat engine is \(34.4\%\). (b) The efficiency of an ideal (Carnot) engine operating between the same two reservoirs is \(64.02\%\).