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Chapter 8: Problem 1202

If a heat engine absorbs \(2 \mathrm{KJ}\) heat from a heat source and release \(1.5 \mathrm{KJ}\) heat into cold reservoir, then its efficiency is (A) \(0.5 \%\) (B) \(75 \%\) (C) \(25 \%\) (D) \(50 \%\)

Short Answer

Expert verified
The efficiency of the heat engine can be calculated using the formula Efficiency = \(\dfrac{Work\ done}{Heat\ absorbed}\) * 100%. First, calculate the work done by subtracting the heat released (1.5 KJ) from the heat absorbed (2 KJ), which gives us 0.5 KJ. Then, apply the formula: Efficiency = \(\dfrac{0.5\ KJ}{2\ KJ}\) * 100% = 25%. Thus, the correct answer is (C).

Step by step solution

01

Identify given values

We are given: - Heat absorbed from heat source: 2 KJ - Heat released into cold reservoir: 1.5 KJ
02

Calculate the work done

To find the work done by the heat engine, we need to subtract the heat released from the heat absorbed. Work done = Heat absorbed - Heat released Work done = 2 KJ - 1.5 KJ Work done = 0.5 KJ
03

Calculate the efficiency

Now we can use the formula for efficiency to calculate the heat engine's efficiency: Efficiency = \(\dfrac{Work\ done}{Heat\ absorbed}\) * 100% Efficiency = \(\dfrac{0.5\ KJ}{2\ KJ}\) * 100% Efficiency = 0.25 * 100% Efficiency = 25% So the efficiency of the heat engine is 25%. The correct answer is (C).

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