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Chapter 8: Problem 1188

Efficiency of a Carnot engine is \(50 \%\), when temperature of outlet is $500 \mathrm{~K}\(. in order to increase efficiency up to \)60 \%$ keeping temperature of intake the same what is temperature of out let. (A) \(200 \mathrm{~K}\) (B) \(400 \mathrm{~K}\) (C) \(600 \mathrm{~K}\) (D) \(800 \mathrm{~K}\)

Short Answer

Expert verified
The new temperature of the outlet to have an efficiency of 60% is 400 K, so the answer is (B) \(400\,\mathrm{K}\).

Step by step solution

01

Write down the Carnot efficiency formula

The efficiency of a Carnot engine is given by: \( \eta = 1 - \frac{T_{outlet}}{T_{intake}} \) Where: - \( \eta \) is the efficiency of the engine - \( T_{outlet} \) is the temperature of the outlet - \( T_{intake} \) is the temperature of the intake
02

Find the temperature of the intake using the given efficiency and outlet temperature

We know the initial efficiency is 50% and the outlet temperature is 500 K. We must find the temperature of the intake using the formula above. \( 0.50 = 1 - \frac{500}{T_{intake}} \) First, we will solve for \( T_{intake} \): \( 0.50 = 1 - \frac{500}{T_{intake}} \Rightarrow T_{intake} = \frac{500}{0.50}\) \( T_{intake} = 1000\,K \)
03

Find the new outlet temperature for the desired efficiency

Now that we have the intake temperature, we want to find the new outlet temperature for an efficiency of 60%. We will use the efficiency formula again with the new efficiency value: \( 0.60 = 1 - \frac{T_{outlet_{new}}}{1000} \) Now solve for \( T_{outlet_{new}} \): \( 0.60 = 1 - \frac{T_{outlet_{new}}}{1000} \Rightarrow T_{outlet_{new}} = 1000 - (1000\times0.60) \) \( T_{outlet_{new}} = 1000 - 600 \) \( T_{outlet_{new}} = 400\,K \)
04

Choose the correct answer from the given options

The new temperature of the outlet to have an efficiency of 60% is 400 K, so the answer is (B) \(400\,\mathrm{K}\).

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Most popular questions from this chapter

70 calorie of heat are required to raise the temperature of 2 mole of an ideal gas at constant pressure from \(30^{\circ} \mathrm{C}\) to $35^{\circ} \mathrm{C}$ The amount of heat required to raise the temperature of the same gas through the same range at constant volume is $\ldots \ldots \ldots \ldots \ldots .$ calorie. (A) 50 (B) 30 (C) 70 (D) 90

A gas mixture consists of 2 mole of oxygen and 4 mole of argon at temperature \(\mathrm{T}\). Neglecting all vibrational modes, the total internal energy of the system is (A) \(11 \mathrm{RT}\) (B) \(9 \mathrm{RT}\) (C) \(15 \mathrm{RT}\) (D) \(4 \mathrm{RT}\)

Which of the following is not a thermodynamical function. (A) Enthalpy (B) Work done (C) Gibb's energy (D) Internal energy

Two cylinders \(\mathrm{A}\) and \(\mathrm{B}\) fitted with piston contain equal amounts of an ideal diatomic gas at \(300 \mathrm{k}\). The piston of \(\mathrm{A}\) is free to move, While that of \(B\) is held fixed. The same amount of heat is given to the gas in each cylinders. If the rise in temperature of the gas in \(\mathrm{A}\) is \(30 \mathrm{~K}\), then the rise in temperature of the gas in \(\mathrm{B}\) is. (A) \(30 \mathrm{~K}\) (B) \(42 \mathrm{~K}\) (C) \(18 \mathrm{~K}\) (D) \(50 \mathrm{~K}\)

One mole of an ideal gas $\left(\mathrm{C}_{\mathrm{p}} / \mathrm{C}_{\mathrm{v}}\right)=\gamma$ at absolute temperature \(\mathrm{T}_{1}\) is adiabatically compressed from an initial pressure \(\mathrm{P}_{1}\) to a final pressure \(\mathrm{P}_{2}\) The resulting temperature \(\mathrm{T}_{2}\) of the gas is given by. (A) $\mathrm{T}_{2}=\mathrm{T}_{1}\left\\{\mathrm{p}_{2} / \mathrm{p}_{1}\right\\}^{\\{\gamma /(\gamma-1)\\}}$ (B) $\mathrm{T}_{2}=\mathrm{T}_{1}\left\\{\mathrm{p}_{2} / \mathrm{p}_{1}\right\\}^{\\{(\gamma-1) / \gamma\\}}$ (C) $\mathrm{T}_{2}=\mathrm{T}_{1}\left\\{\mathrm{p}_{2} / \mathrm{p}_{1}\right\\}^{\gamma}$ (D) $\mathrm{T}_{2}=\mathrm{T}_{1}\left(\mathrm{p}_{2} / \mathrm{p}_{1}\right)^{\gamma-1}$
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