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 :
\[\eta =\frac{W}{Q}=1-\frac{{{T}_{2}}}{{{T}_{1}}}\] Work, \[W=\frac{{{T}_{1}}-{{T}_{2}}}{{{T}_{1}}}{{Q}_{1}}\] \[W=\frac{(227+273)-(127+273)}{(227+273)}\times 6.0\times {{10}^{4}}\text{cal}\] \[=\frac{100}{500}\times 6.0\times {{10}^{4}}\text{cal}\] \[=1.2\times {{10}^{4}}\text{cal}\]
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