Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

Carnot Cycle and Efficiency

An engine operates on a Carnot cycle between a hot reservoir at $\mathrm{600 K}$ and a cold reservoir at $\mathrm{300 K}$ . Calculate the efficiency of the engine and the heat absorbed from the hot reservoir if it expels $\mathrm{4000 J }$ of heat to the cold reservoir.

Option: 1
$\mathrm{8000J}$

Option: 2
$\mathrm{4690 J}$

Option: 3
$\mathrm{2520 J}$

Option: 4
$\mathrm{1000J}$

Answers (1)

best_answer

Given data:

Temperature of hot reservoir, $\mathrm{T_h = 600 K}$

Temperature of cold reservoir, $\mathrm{T_c = 300 K}$

Heat expelled to cold reservoir, $\mathrm{Q_c = 4000 K}$

The efficiency of a Carnot engine is given by the formula:

Efficiency $\mathrm{=1-\frac{T_c}{T_h} }$

Efficiency $\mathrm{=1-\frac{300 \mathrm{~K}}{600 \mathrm{~K}} }$

Efficiency $\mathrm{=0.5}$

The heat absorbed from the hot reservoir can be calculated using the efficiency formula:

Efficiency $\mathrm{=\frac{Q_h-Q_c}{Q_h}}$

Where $\mathrm{Q_h}$ is the heat absorbed from the hot reservoir.

Solving for $\mathrm{Q_h :}$

$\mathrm{0.5 =\frac{Q_h-4000 \mathrm{~J}}{Q_h} }$

$\mathrm{0.5 Q_h =Q_h-4000 \mathrm{~J} }$

$\mathrm{0.5 Q_h =4000 \mathrm{~J} }$

$\mathrm{Q_h =8000 \mathrm{~J} }$

Therefore,the correct option is 1

Posted by

qnaprep

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

50 mL of 0.2 M ammonia solution is treated with 25 mL of 0.2 M HCl. If pK_b of ammonia solution is 4.75, the pH of the mixture will be :
Option: 1 3.75
Option: 2 4.75

Trending Articles/News

anam-khan

JEE Main 2026: JCECEB invites fresh applications for Jharkhand state merit list correction

Latest Question