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Chapter 8: Problem 1199

The efficiency of Carnot's engine operating between reservoirs, maintained at temperature \(27^{\circ} \mathrm{C}\) and \(-123^{\circ} \mathrm{C}\) is (A) \(0.5\) (B) \(0.4\) (C) \(0.6\) (D) \(0.25\)

Short Answer

Expert verified
The efficiency of the Carnot engine operating between reservoirs maintained at temperatures \(27^{\circ} \mathrm{C}\) and \(-123^{\circ} \mathrm{C}\) is \(0.5\) (Option A). This is determined by firstly converting the given Celsius temperatures to Kelvin (\(T_h = 300 K\) and \(T_c = 150 K\)), then using the Carnot efficiency formula (\(\eta = 1 - \frac{T_{c}}{T_{h}}\)) to calculate the efficiency.

Step by step solution

01

Convert temperatures to Kelvin

Given temperatures in Celsius are \(27^\circ C\) and \(-123^\circ C\). To convert them to Kelvin, just add 273 to the Celsius values. Hot reservoir temperature: \(T_h = 27^\circ C + 273 = 300 K\) Cold reservoir temperature: \(T_c = -123^\circ C + 273 = 150 K\)
02

Calculate Efficiency

Now that we have the temperatures in Kelvin, we can use the Carnot efficiency formula to find the efficiency of the engine: \(\eta = 1 - \frac{T_{c}}{T_{h}} = 1 - \frac{150}{300}\) Now, simplify the fraction inside: \(\eta = 1 - \frac{1}{2}\) Finally, calculate the efficiency; \(\eta = 1 - 0.5 = 0.5\)
03

Select the correct answer

Our calculated efficiency, \( \eta = 0.5\), corresponds to option (A). So, the correct answer is: (A) \(0.5\)

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Most popular questions from this chapter

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