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Chapter 15: Problem 33

(a) Calculate the efficiency of a reversible engine that operates between the temperatures \(600.0^{\circ} \mathrm{C}\) and \(300.0^{\circ} \mathrm{C} .\) (b) If the engine absorbs \(420.0 \mathrm{kJ}\) of heat from the hot reservoir, how much does it exhaust to the cold reservoir?

Short Answer

Expert verified
Answer: The engine exhausts 275.86 kJ of heat to the cold reservoir.

Step by step solution

01

Convert temperatures to Kelvin

To use the efficiency formula, we first need to convert the given Celsius temperatures to Kelvin. To do so, simply add 273.15 to the Celsius temperature: \(T_{hot} = 600.0 + 273.15 = 873.15 \mathrm{K}\) \(T_{cold} = 300.0 + 273.15 = 573.15 \mathrm{K}\)
02

Calculate the efficiency of the engine

We can use the formula for the efficiency of a reversible heat engine: Efficiency = \(1 - \frac{T_{cold}}{T_{hot}}\) Plug in the temperatures in Kelvin: Efficiency = \(1 - \frac{573.15}{873.15} = 1 - 0.6568 = 0.3432 = 34.32\%\)
03

Calculate the heat exhausted to the cold reservoir

Now that we have the efficiency, we can use the formula to find the heat exhausted to the cold reservoir: \(Q_{cold} = Q_{hot} * (1 - \text{Efficiency})\) Given that the engine absorbs \(420.0 \mathrm{kJ}\) of heat from the hot reservoir: \(Q_{cold} = 420.0 * (1 - 0.3432) = 420.0 * 0.6568 = 275.86 \mathrm{kJ}\) So, the engine exhausts \(275.86 \mathrm{kJ}\) of heat to the cold reservoir.

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