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Chapter 8: Problem 1194

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) \(99^{\circ} \mathrm{C}, 37^{\circ} \mathrm{C}\)

Short Answer

Expert verified
The temperatures of the source and sink are \(99^{\circ}\mathrm{C}\) and \(37^{\circ}\mathrm{C}\), which corresponds to option (D).

Step by step solution

01

Set up the initial efficiency equation

Using the efficiency formula, we have \( \frac{1}{6} = 1 - \frac{T_s}{T_r} \) Where \(T_s\) is the temperature of the sink, and \(T_r\) is the temperature of the source.
02

Set up the efficiency equation after the sink temperature is reduced

When the sink temperature is reduced by 62°C, the efficiency doubles. Let the new sink temperature be marked as \(T_s' = T_s - 62 \). The efficiency equation becomes: \( \frac{1}{3} = 1 - \frac{T_s - 62}{T_r} \)
03

Solve the system of equations

Now, we have a system of two equations: 1. \( \frac{1}{6} = 1 - \frac{T_s}{T_r} \) 2. \( \frac{1}{3} = 1 - \frac{T_s - 62}{T_r} \) To solve this, we can rewrite equation (1): \(T_s = T_r - \frac{T_r}{6} \) Now substitute this in equation (2): \( \frac{1}{3} = 1 - \frac{(T_r - \frac{T_r}{6}) - 62}{T_r} \) Solve for \(T_r\).
04

Calculate the temperature of the source

After solving the equation, we find that the temperature of the source is: \( T_r = 372 K \) We need to convert this to degrees Celsius: \[ T_r^{\circ}C = T_rK - 273 \] \( T_r^{\circ}C = 372 - 273 = 99^{\circ}C \)
05

Calculate the temperature of the sink

Now we can substitute the source temperature back into equation (1) to get the sink temperature: \( T_s = T_r - \frac{T_r}{6} \) \( T_s = 99 - \frac{99}{6} = 310 K \) To convert this to degrees Celsius: \[ T_s^{\circ}C = T_sK - 273 \] \( T_s^{\circ}C = 310 - 273 = 37^{\circ}C \) This matches option (D) \(99^{\circ}\mathrm{C}, 37^{\circ}\mathrm{C}\).

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Most popular questions from this chapter

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) \(25 \%\)

One \(\mathrm{kg}\) of adiatomic gas is at a pressure of $5 \times 10^{5}\left(\mathrm{~N} / \mathrm{m}^{2}\right)$ The density of the gas is \(\left\\{(5 \mathrm{~kg}) / \mathrm{m}^{3}\right\\}\) what is the energy of the gas due to its thermal motion ? (A) \(2.5 \times 10^{5} \mathrm{~J}\) (B) \(3.5 \times 10^{5} \mathrm{~J}\) (C) \(4.5 \times 10^{5} \mathrm{~J}\) (D) \(1.5 \times 10^{5} \mathrm{~J}\)

A System under goes a Cyclic Process in which it absorbs \(\mathrm{Q}_{1}\) heat and gives out \(\mathrm{Q}_{2}\) heat. The efficiency of the Process \(\eta\) and the work done is \(\mathrm{W}\). Which formula is wrong ? is (A) \(\mathrm{W}=\mathrm{Q}_{1}-\mathrm{Q}_{2}\) (B) \(\eta=\left(\mathrm{Q}_{2} / \mathrm{Q}_{1}\right)\) (C) \(\eta=\left(\mathrm{W} / \mathrm{Q}_{1}\right)\) (D) \(\eta=1-\left(\mathrm{Q}_{2} / \mathrm{Q}_{1}\right)\)

Two identical samples of a gas are allowed to expand (i) isothermally (ii) adiabatically work done is (A) more in an isothermal process (B) more in an adiabatic process (C) equal in both process. (D) neither of them

Instructions:Read the assertion and reason carefully to mask the correct option out of the options given below. (A) If both assertion and reason are true and the reason is the correct explanation of the assertion. (B) If both assertion and reason are true but reason is not be correct explanation of assertion. (C) If assertion is true but reason is false. (D) If the assertion and reason both are false. Assertion: The total translation kinetic energy of all the molecules of a given mass of an ideal gas is \(1.5\) times the product of its Pressure and its volume. Reason: The molecules of a gas collide with each other and velocities of the molecules change due to the collision (A) D (B) \(\mathrm{C}\) (C) A (D) B
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