JEE PYQ: Thermodynamics Question 27
Question 27 - 2020 (07 Jan Shift 1)
A Carnot engine operates between two reservoirs of temperatures 900 K and 300 K. The engine performs 1200 J of work per cycle. The heat energy (in J) delivered by the engine to the low temperature reservoir, in a cycle, is __________.
Show Answer
Answer: 600
Solution
Given: $T_1 = 900$ K, $T_2 = 300$ K, $W = 1200$ J
Using, $1 - \frac{T_2}{T_1} = \frac{W}{Q_1}$
$1 - \frac{300}{900} = \frac{1200}{Q_1}$
$\frac{2}{3} = \frac{1200}{Q_1} \Rightarrow Q_1 = 1800$
Therefore heat energy delivered by the engine to the low temperature reservoir, $Q_2 = Q_1 - W = 1800 - 1200 = 600$ J
BETA
