JEE PYQ: Thermodynamics Question 14
Question 14 - 2020 (02 Sep Shift 2)
A heat engine is involved with exchange of heat of 1915 J, $-40$ J, $+125$ J and $-Q$ J, during one cycle achieving an efficiency of 50.0%. The value of $Q$ is :
(1) 640 J
(2) 40 J
(3) 980 J
(4) 400 J
Show Answer
Answer: (3)
Solution
Efficiency, $\eta = \frac{\text{Work done}}{\text{Heat absorbed}} = \frac{W}{\Sigma Q} = 0.5$
$= \frac{Q_1 + Q_2 + Q_3 + Q_4}{Q_1 + Q_3} = 0.5$
Here, $Q_1 = 1915$ J, $Q_2 = -40$ J and $Q_3 = 125$ J
$\frac{1915 - 40 + 125 + Q_4}{1915 + 125} = 0.5$
$1915 - 40 + 125 + Q_4 = 1020$
$Q_4 = 1020 - 2000$
$Q_4 = -Q = -980$ J $\Rightarrow Q = 980$ J
BETA
