A) \[4.8\times {{10}^{4}}\text{ }cal\]
B) \[1.2\times {{10}^{4}}\text{ }cal\]
C) \[3.5\times {{10}^{4}}\text{ }cal\]
D) \[1.6\times {{10}^{4}}\text{ }cal\]
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, \[T={{227}^{o}}C=227+273=500K,\] \[{{T}_{2}}={{127}^{o}}C=127+273=400K\] \[\therefore \] \[W=\left( 1-\frac{400}{500} \right)(6\times {{10}^{4}})\] \[W=\frac{6\times {{10}^{4}}}{5}=1.2\times {{10}^{4}}cal\]
You need to login to perform this action.
You will be redirected in
3 sec
