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 :
From the relation \[\frac{{{Q}_{2}}}{{{Q}_{1}}}=\frac{{{T}_{2}}}{{{T}_{1}}}\] Given, \[{{Q}_{1}}=6\times {{10}^{4}}\] cal, \[{{T}_{1}}=227+273=500\,K\], \[{{T}_{2}}=127+273=400\,K\]. \[\therefore \] \[\frac{{{Q}_{2}}}{6\times {{10}^{4}}}=\frac{400}{500}\] \[\Rightarrow \] \[{{Q}_{2}}=\frac{4}{5}\times 6\times {{10}^{4}}\] \[=4.8\times {{10}^{4}}\] cal Now, heat converted to work \[={{Q}_{1}}-{{Q}_{2}}\] \[=6.0\times {{10}^{4}}-4.8\times {{10}^{4}}\] \[=1.2\times {{10}^{4}}\] cal Note: Carnot cycle consists of following four stages : (i) Isothermal expansion (ii) Adiabatic expansion (iii) Isothermal compression (iv) Adiabatic compression After doing the calculations for different processes, we achieve the relation \[\frac{{{Q}_{2}}}{{{Q}_{1}}}=\frac{{{T}_{2}}}{{{T}_{1}}}\]
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