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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

\(\mu\) moles of gas expands from volume \(\mathrm{V}_{1}\) to \(\mathrm{V}_{2}\) at constant temperature \(\mathrm{T}\). The work done by the gas is (A) \(\mu \mathrm{RT}\left\\{\mathrm{V}_{2} / \mathrm{V}_{1}\right\\}\) (B) \(\mu \operatorname{RTln}\left\\{\mathrm{V}_{2} / \mathrm{V}_{1}\right\\}\) (C) $\mu \mathrm{RT}\left\\{\left(\mathrm{V}_{\mathrm{v}} / \mathrm{V}_{1}\right)-1\right\\}$ (D) $\mu \operatorname{RTln}\left\\{\left(\mathrm{V}_{2} / \mathrm{V}_{1}\right)\right.$

A diatomic gas initially at \(18^{\circ} \mathrm{C}\) is Compressed adiabatically to one eighth of its original volume. The temperature after Compression will be (A) \(10^{\circ} \mathrm{C}\) (B) \(668 \mathrm{~K}\) (C) \(887^{\circ} \mathrm{C}\) (D) \(144^{\circ} \mathrm{C}\)

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) \(25 \%\)

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.

For an adiabatic process involving an ideal gas (A) \(\mathrm{P}^{\gamma-1}=\mathrm{T}^{\gamma-1}=\) constant (B) \(\mathrm{P}^{1-\gamma}=\mathrm{T}^{\gamma}=\) constant (C) \(\mathrm{PT}^{\gamma-1}=\) constant (D) \(\mathrm{P}^{\gamma-1}=\mathrm{T}^{\gamma}=\) constant
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