A Carnot engine operating between temperatures $600 \mathrm{~K}$ and $300 \mathrm{~K}$, absorbs $800…

A Carnot engine operating between temperatures $600 \mathrm{~K}$ and $300 \mathrm{~K}$, absorbs $800 \mathrm{~J}$ of heat from the source. The mechanical work done per cycle is
  1. $400 \mathrm{~J}$
  2. $650 \mathrm{~J}$
  3. $750 \mathrm{~J}$
  4. $600 \mathrm{~J}$

Solution

$\eta=1-\frac{300}{600}=1-\frac{1}{2}=\frac{1}{2}$ Now, $w=\eta Q=\frac{1}{2} \times 800=400 J$

Asked in: AP EAMCET 2022 (05 Jul Shift 2)

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