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Chapter 8: Problem 1193

An engine is supposed to operate between two reservoirs at temperature \(727^{\circ} \mathrm{C}\) and \(227^{\circ} \mathrm{C}\). The maximum possible efficiency of such an engine is (A) \((3 / 4)\) (B) \((1 / 4)\) (C) \((1 / 2)\) (D) 1

Short Answer

Expert verified
The maximum possible efficiency of an engine operating between two reservoirs at \(727^{\circ} \mathrm{C}\) and \(227^{\circ} \mathrm{C}\) is computed by first converting the temperatures to Kelvin giving \(1000.15 K\) and \(500.15 K\) respectively. The efficiency of a Carnot engine is determined by the formula \(Efficiency = 1 - \frac{T_C}{T_H}\). Substituting the converted temperatures into the formula results to \(Efficiency = 1 - \frac{500.15}{1000.15} = 0.5\). Hence, the answer is (C) \(\frac{1}{2}\).

Step by step solution

01

Convert temperatures to Kelvin

Since the given temperatures are in Celsius, we first need to convert them to Kelvin by adding 273.15 to each temperature: Hot reservoir: \( T_H = 727 + 273.15 = 1000.15 K \) Cold reservoir: \( T_C = 227 + 273.15 = 500.15 K \)
02

Determine the maximum efficiency

Now that we have temperatures in Kelvin, we can use the formula for the maximum efficiency of a Carnot engine, which is given by: \(Efficiency = 1 - \frac{T_C}{T_H}\)
03

Calculate the maximum efficiency

We can now plug in the values for the hot and cold reservoir temperatures into the efficiency formula: \(Efficiency = 1 - \frac{500.15}{1000.15}\)
04

Simplify the result

After we plug the values into the formula, we can calculate the efficiency: \(Efficiency = 1 - 0.5 = 0.5\)
05

Choose the correct answer

Comparing our result to the answer choices given, we can see that the maximum possible efficiency of the engine is (C) \(\frac{1}{2}\).

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Most popular questions from this chapter

A gas expands \(0.25 \mathrm{~m}^{3}\) at Constant Pressure \(10^{3}\left(\mathrm{~N} / \mathrm{m}^{2}\right)\) the work done is (A) \(250 \mathrm{~J}\) (B) \(2.5\) erg (C) \(250 \mathrm{~W}\) (D) \(250 \mathrm{~N}\)

A Carnot engine Converts one sixth of the heat input into work. When the temperature of the sink is reduced by \(62^{\circ} \mathrm{C}\) the efficiency of the engine is doubled. The temperature of the source and sink are (A) \(80^{\circ} \mathrm{C}, 37^{\circ} \mathrm{C}\) (B) \(95^{\circ} \mathrm{C}, 28^{\circ} \mathrm{C}\) (C) \(90^{\circ} \mathrm{C}, 37^{\circ} \mathrm{C}\) (D) \(99^{\circ} \mathrm{C}, 37^{\circ} \mathrm{C}\)

For which combination of working temperatures the efficiency of Carnot's engine is highest. (A) \(80 \mathrm{~K}, 60 \mathrm{~K}\) (B) \(100 \mathrm{~K}, 80 \mathrm{~K}\) (C) \(60 \mathrm{~K}, 40 \mathrm{~K}\) (D) \(40 \mathrm{~K}, 20 \mathrm{~K}\)

An ideal refrigerator has a freezer at a temperature of \(-13\) C, The coefficient of performance of the engine is 5 . The temperature of the air to which heat is rejected will be. (A) \(325^{\circ} \mathrm{C}\) (B) \(39^{\circ} \mathrm{C}\) (C) \(325 \mathrm{~K}\) (D) \(320^{\circ} \mathrm{C}\)

If the ratio of specific heat of a gas at Constant pressure to that at constant volume is \(\gamma\), the Change in internal energy of the mass of gas, when the volume changes from \(\mathrm{V}\) to \(2 \mathrm{~V}\) at Constant Pressure p, is (A) \(\\{(\mathrm{PV}) /(\gamma-1)\\}\) (B) \(\\{\mathrm{R} /(\gamma-1)\\}\) (C) PV (D) \(\\{(\gamma \mathrm{PV}) /(\gamma-1)\\}\)
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