Thermodynamics Question 145
Question: A Carnot engine working between $ 300K $ and 600K has work output of 800 J per cycle. What is amount of heat energy supplied to the engine from source per cycle
[DPMT 1999; Pb. PMT 2002, 05; Kerala PMT 2004]
Options:
A) 1800 J/cycle
B) 1000 J/cycle
C) 2000 J/cycle
D) 1600 J/cycle
Show Answer
Answer:
Correct Answer: D
Solution:
$ \eta =\frac{T _{1}-T _{2}}{T _{1}}-\frac{W}{Q} $
Therefore $ Q=( \frac{T _{1}}{T _{1}-T _{2}} )W $
$ =\frac{600}{(600-300)}\times 800 $ =1600 J
BETA
