JEE Main & Advanced Physics Thermodynamical Processes Question Bank Topic Test - Thermodynamical Process (2021)

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

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

Correct Answer: C

Solution :

[c] \[\eta =\frac{{{T}_{1}}-{{T}_{2}}}{{{T}_{1}}}=\frac{W}{Q}\]Þ \[W=\frac{Q({{T}_{1}}-{{T}_{2}})}{{{T}_{1}}}\]

\[=\frac{6\times {{10}^{4}}\left[ (227+273)-(273+127) \right]}{(227+273)}\]

\[=\frac{6\times {{10}^{4}}\times 100}{500}\]\[=1.2\times {{10}^{4}}cal\]