A) \[4.8\times {{10}^{4}}cal\]
B) \[1.2\times {{10}^{4}}cal\]
C) \[3.5\times {{10}^{4}}cal\]
D) \[1.6\times {{10}^{4}}cal\]
Correct Answer: B
Solution :
For a heat engine the efficiency is the ratio of useful work performed to the heat energy consumed from the high temperature reservoir \[\eta =\frac{W}{Q}\] \[\Rightarrow \] \[W=\eta Q\] ...(i) Also, \[\eta =1-\frac{{{T}_{2}}}{{{T}_{1}}}\] ...(ii) From Eqs. (i) and (ii), we have \[W=\left( 1-\frac{{{T}_{2}}}{{{T}_{1}}} \right).Q\] Given, \[T=227{}^\circ C=227+273=500\text{ }K,\] \[{{T}_{2}}=127{}^\circ C=127+273=400\text{ }K\] \[\therefore \] \[W=\left( 1-\frac{400}{500} \right)(6\times {{10}^{4}})\] \[W=\frac{6\times {{10}^{4}}}{5}=1.2\times {{10}^{4}}\,cal\]
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