Thermodynamics Question 356

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

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

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Answer:

Correct Answer: C

Solution:

[c] $ \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 $



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