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 :
we have \[\frac{Q}{T}=\text{constant}\Rightarrow \frac{{{Q}_{2}}}{{{Q}_{1}}}=\frac{{{T}_{2}}}{{{T}_{1}}}\] 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 \[{{Q}_{1}}={{Q}_{2}}=6.0\times {{10}^{4}}-4.8\times {{10}^{4}}=1.2\times {{10}^{4}}cal\] Aliter:\[\eta =\frac{{{T}_{1}}-{{T}_{2}}}{{{T}_{1}}}=\frac{W}{Q}\Rightarrow W=\frac{Q({{T}_{1}}-{{T}_{2}})}{{{T}_{1}}}\] \[W=\frac{6\times {{10}^{4}}[(227+273)-(237+127)]}{(227+273)}\] \[\Rightarrow \]\[W=\frac{6\times {{10}^{4}}\times 100}{500}=1.2\times {{10}^{4}}\,cal\]
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