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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

In a container of negligible heat capacity, \(200 \mathrm{~g}\) ice at \(0^{\circ} \mathrm{C}\) and \(100 \mathrm{~g}\) steam at \(100^{\circ} \mathrm{C}\) are added to \(200 \mathrm{~g}\) of water that has temperature $55^{\circ} \mathrm{C}$. Assume no heat is lost to the surroundings and the pressure in the container is constant atm. What is the final temperature the System? (A) \(72^{\circ} \mathrm{C}\) (B) \(48^{\circ} \mathrm{C}\) (C) \(100^{\circ} \mathrm{C}\) (D) \(94^{\circ} \mathrm{C}\)

A monoatomic gas is used in a Carnot engine as the working substance, If during the adiabatic expansion part of the cycle the volume of the gas increases from \(\mathrm{V}\) to \(8 \mathrm{~V}_{1}\) the efficiency of the engine is.. (A) \(60 \%\) (B) \(50 \%\) (C) \(75 \%\) (D) \(25 \%\)

For hydrogen gas \(\mathrm{C}_{\mathrm{p}}-\mathrm{C}_{\mathrm{v}}=\mathrm{a}\) and for oxygen gas \(\mathrm{C}_{\mathrm{p}}-\mathrm{C}_{\mathrm{v}}=\mathrm{b}\), The relation between \(\mathrm{a}\) and \(\mathrm{b}\) is given by (A) \(a=4 b\) (B) \(a=b\) (C) \(\mathrm{a}=16 \mathrm{~b}\) (D) \(a=8 b\)

The latent heat of Vaporization of water is \(2240(\mathrm{~J} / \mathrm{g})\) If the work done in the Process of expansion of \(1 \mathrm{~g}\) is $168 \mathrm{~J}$. then increase in internal energy is ......... J (A) 2072 (B) 2408 (C) 2240 (D) 1904

In a container of negligible heat capacity, \(200 \mathrm{~g}\) ice at \(0^{\circ} \mathrm{C}\) and \(100 \mathrm{~g}\) steam at \(100^{\circ} \mathrm{C}\) are added to \(200 \mathrm{~g}\) of water that has temperature $55^{\circ} \mathrm{C}$. Assume no heat is lost to the surroundings and the pressure in the container is constant \(1 \mathrm{~atm} .\) Amount of the Sm left in the system, is equal to (A) \(16.7 \mathrm{~g}\) (B) \(8.4 \mathrm{~g}\) (C) \(12 \mathrm{~g}\) (D) \(0 \mathrm{~g}\) Copyright () StemEZ.com. All rights reserved.
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