Search any question & find its solution
Question:
Answered & Verified by Expert
Carnot heat engine takes $300 \mathrm{~J}$ of heat from a source at $627^{\circ} \mathrm{C}$ and gives some part of it to $\operatorname{sink}$ at $27^{\circ} \mathrm{C}$. Work done by engine in one cycle is
Options:
Solution:
1602 Upvotes
Verified Answer
The correct answer is:
$200 \mathrm{~J}$
The efficiency of a Carnot engine is given by
$\eta=1-\frac{T_{2}}{T_{1}}$
Given, $\quad T_{1}=627^{\circ} \mathrm{C}=627+273=900 \mathrm{~K}$
and $\quad T_{2}=27^{\circ} \mathrm{C}=27+273=300 \mathrm{~K}$
$\Rightarrow \quad \eta=1-\frac{300}{900}=\frac{600}{900}=\frac{2}{3}$
Also, $\quad \eta=\frac{\text { Work done }}{\text { Heat taken }}$
$\Rightarrow$ Work done $=\eta \times$ Heat taken
$=\frac{2}{3} \times 300=200 \mathrm{~J}$
$\eta=1-\frac{T_{2}}{T_{1}}$
Given, $\quad T_{1}=627^{\circ} \mathrm{C}=627+273=900 \mathrm{~K}$
and $\quad T_{2}=27^{\circ} \mathrm{C}=27+273=300 \mathrm{~K}$
$\Rightarrow \quad \eta=1-\frac{300}{900}=\frac{600}{900}=\frac{2}{3}$
Also, $\quad \eta=\frac{\text { Work done }}{\text { Heat taken }}$
$\Rightarrow$ Work done $=\eta \times$ Heat taken
$=\frac{2}{3} \times 300=200 \mathrm{~J}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.