Thermodynamics Question 171
Question: An ideal gas heat engine operates in Carnot cycle between 227°C and 127°C. It absorbs $ 6\times 10^{4} $ cals of heat at higher temperature. Amount of heat converted to work is
[CBSE PMT 2005]
Options:
A) $ 2.4\times 10^{4} $ cal
B) $ 6\times 10^{4} $ cal
C) $ 1.2\times 10^{4} $ cal
D) $ 4.8\times 10^{4} $ cal
Show Answer
Answer:
Correct Answer: C
Solution:
$ \eta =\frac{T _{1}-T _{2}}{T _{1}}=\frac{W}{Q} $
Therefore $ W=\frac{Q(T _{1}-T _{2})}{T _{1}} $
$ =\frac{6\times 10^{4}[ (227+273)-(273+127) ]}{(227+273)} $
$ =\frac{6\times 10^{4}\times 100}{500} $
$ =1.2\times 10^{4}cal $
