Search any question & find its solution
Question:
Answered & Verified by Expert
A Carnot engine takes $3 \times 10^6 \mathrm{cal}$. of heat from a reservoir at $627^{\circ} \mathrm{C}$, and gives it to a sink at $27^{\circ} \mathrm{C}$. The work done by the engine is
Options:
Solution:
2371 Upvotes
Verified Answer
The correct answer is:
$8.4 \times 10^6 \mathrm{~J}$
$8.4 \times 10^6 \mathrm{~J}$
$\eta=\frac{(627+273)-(273+27)}{627+273}$
$=\frac{900-300}{900}=\frac{600}{900}=\frac{2}{3}$
work $=(\eta) \times$ Heat
$=\frac{2}{3} \times 3 \times 10^6 \times 4.2 \mathrm{~J}=8.4 \times 10^6 \mathrm{~J}$
$=\frac{900-300}{900}=\frac{600}{900}=\frac{2}{3}$
work $=(\eta) \times$ Heat
$=\frac{2}{3} \times 3 \times 10^6 \times 4.2 \mathrm{~J}=8.4 \times 10^6 \mathrm{~J}$
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.