JEE PYQ: Thermodynamics Question 2
Question 2 - 2021 (17 Mar Shift 1)
A Carnot’s engine working between 400 K and 800 K has a work output of 1200 J per cycle. The amount of heat energy supplied to the engine from the source in each cycle is :
(1) 3200 J
(2) 1800 J
(3) 1600 J
(4) 2400 J
Show Answer
Answer: (4)
Solution
$\eta = \frac{T_2}{T_1} = \frac{Q_2}{Q_1} = \frac{Q_1 - W}{Q_1}$ $(\because W = Q_1 - Q_2)$
$\frac{400}{800} = 1 - \frac{W}{Q_1}$
$\frac{W}{Q_1} = 1 - \frac{1}{2} = \frac{1}{2}$
$Q_1 = 2W = 2400$ J
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