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Chapter 8: Problem 1195

Carnot engine working between \(300 \mathrm{~K}\) and \(600 \mathrm{~K}\) has work output of \(800 \mathrm{~J}\) per cycle. What is amount of heat energy supplied to the engine from source per cycle (A) \(1600\\{\mathrm{~J} /\) (cycle)\\} (B) \(2000\\{\mathrm{~J} /(\mathrm{cycle})\\}\) (C) \(1000\\{\mathrm{~J} /(\) cycle \()\\}\) (D) \(1800\\{\mathrm{~J} /(\) cycle \()\\}\)

Short Answer

Expert verified
The amount of heat energy supplied to the engine from the source per cycle is 1600 J. So, the correct answer is (A) \(1600 \mathrm{~J / cycle}\).

Step by step solution

01

Calculate the Efficiency of Carnot Engine

First, let's find out the efficiency of this Carnot engine. The efficiency of a Carnot engine depends on the hot and cold temperatures: \[η_{Carnot} = 1 - \frac{T_{cold}}{T_{hot}}\] where, \(η_{Carnot}\) is the efficiency \(T_{hot}\) = 600 K (temperature of heat source) \(T_{cold}\) = 300 K (temperature of heat sink) Substituting the given values in the formula, we get: \[η_{Carnot} = 1 - \frac{300}{600}\]
02

Solve for Efficiency

Now, we will solve the expression to find η: \[η_{Carnot} = 1 - \frac{1}{2}\] \[η_{Carnot} = \frac{1}{2}\] So, the efficiency of the Carnot engine is 0.5, or 50%.
03

Use the Efficiency and Work Output to Find the Heat Energy Supplied

We can now use the efficiency and work output to find the heat energy supplied per cycle. The formula for Carnot engine efficiency is: \[η_{Carnot} = \frac{W_{output}}{Q_{input}}\] Where, \(W_{output}\) = work output per cycle (800 J) \(Q_{input}\) = heat energy supplied to the engine per cycle Rearranging the formula to find \(Q_{input}\): \[Q_{input} = \frac{W_{output}}{η_{Carnot}}\] Now, we substitute the given values and the efficiency value calculated in step 2: \[Q_{input} = \frac{800}{0.5}\]
04

Calculate the Heat Energy Supplied

Now we solve the expression to get the heat energy supplied per cycle: \[Q_{input} = 1600 J\] The amount of heat energy supplied to the engine from the source per cycle is 1600 J. So, the correct answer is (A) \(1600 \mathrm{~J / cycle}\).

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Most popular questions from this chapter

\(\mathrm{Cp}\) and Cv denote the specific heat of oxygen per unit mass at constant Pressure and volume respectively, then (A) \(\mathrm{cp}-\mathrm{cv}=(\mathrm{R} / 16)\) (B) \(\mathrm{Cp}-\mathrm{Cv}=\mathrm{R}\) (C) \(\mathrm{Cp}-\mathrm{Cv}=32 \mathrm{R}\) (D) \(\mathrm{Cp}-\mathrm{Cv}=(\mathrm{R} / 32)\)

\(1 \mathrm{~mm}^{3}\) Of a gas is compressed at 1 atmospheric pressure and temperature \(27^{\circ} \mathrm{C}\) to \(627^{\circ} \mathrm{C}\) What is the final pressure under adiabatic condition. \(\mathrm{r}=1.5\) (A) \(80 \times 10^{5}\left(\mathrm{~N} / \mathrm{m}^{2}\right)\) (B) \(36 \times 10^{5}\left(\mathrm{~N} / \mathrm{m}^{2}\right)\) (C) \(56 \times 10^{5}\left(\mathrm{~N} / \mathrm{m}^{2}\right)\) (D) \(27 \times 10^{5}\left(\mathrm{~N} / \mathrm{m}^{2}\right)\)

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) \(\eta=1-\left(\mathrm{Q}_{2} / \mathrm{Q}_{1}\right)\)

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Work done per mole in an isothermal ? change is (A) RT log \(_{e}\left(\mathrm{v}_{2} / \mathrm{v}_{1}\right)\) (B) \(\mathrm{RT} \log _{10}\left(\mathrm{v}_{2} / \mathrm{v}_{1}\right)\) (C) RT \(\log _{10}\left(\mathrm{v}_{1} / \mathrm{v}_{2}\right)\) (D) RT \(\log _{\mathrm{e}}\left(\mathrm{v}_{1} / \mathrm{v}_{2}\right)\)
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