A) \[4.8\times {{10}^{4}}cal\]
B) \[3.5\times {{10}^{4}}cal\]
C) \[1.6\times {{10}^{4}}cal\]
D) \[1.2\times {{10}^{4}}cal\]
Correct Answer: D
Solution :
Efficiency\[(\eta )\]is given by \[\eta =\frac{work\text{ }done}{heat\text{ }absorbed}=1-\frac{{{T}_{1}}}{{{T}_{2}}}\] Given, \[{{T}_{1}}={{127}^{o}}C=273+127=400\text{ }K,\] \[{{T}_{2}}={{227}^{o}}C=227+273=500\text{ }K\] \[\eta =1-\frac{400}{500}=0.2\] \[\Rightarrow \]Work done \[=0.2\times 6\times {{10}^{4}}=1.2\times {{10}^{4}}cal\]
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