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JEE Main 202117 MarchMorning Shift
Exam
JEE Main
Year
2021
Date
17 March
Shift
Morning Shift
Subject
Physics
Chapter
Thermodynamics

JEE Main 2021 · 17 March · Morning Shift · Physics · Thermodynamics

A Carnot's engine working between 400K and 800K has a work output...

A Carnot's engine working between 400K400K and 800K800K has a work output of 1200J1200J per cycle. The amount of heat energy supplied to the engine from the source in each cycle is:

Options

  1. A3200J3200J
  2. B1800J1800J
  3. C1600J1600J
  4. D2400J2400J

Correct answer

D. 2400J2400J

Step-by-step solution

η=T2T1=Q2Q1=Q1−WQ1(∵W=Q1−Q2)η = \dfrac{{T}_{2}}{{T}_{1}} = \dfrac{{Q}_{2}}{{Q}_{1}} = \dfrac{{Q}_{1} - W}{{Q}_{1}}\left(∵ W = {Q}_{1} - {Q}_{2}\right)

400800=1−WQ1\dfrac{400}{800} = 1 - \dfrac{W}{{Q}_{1}}

WQ1=1−12=12\dfrac{W}{{Q}_{1}} = 1 - \dfrac{1}{2} = \dfrac{1}{2}

Q1=2W=2400J{Q}_{1} = 2W = 2400J

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