A) 325 K
B) 375 K
C) 300 K
D) 350 K
Correct Answer: C
Solution :
The efficiency of Carnot engine is defined as the ratio of useful work obtained from the engine to the heat supplied to it. i.e., \[\eta =work\frac{output}{heat\text{ }input}=\frac{W}{{{Q}_{H}}}\] or \[\eta =\frac{{{Q}_{H}}-{{Q}_{L}}}{{{Q}_{H}}}=1-\frac{{{Q}_{L}}}{{{Q}_{H}}}\] Also, we can show that \[\frac{{{Q}_{L}}}{{{Q}_{H}}}=\frac{{{T}_{L}}}{{{T}_{H}}}\] \[\therefore \] \[\eta =1-\frac{{{T}_{L}}}{{{T}_{H}}}\] where\[{{T}_{L}}\]is temperature of sink and\[{{T}_{H}}\]is temperature of hot reservoir. According to question. \[\frac{1}{5}=1-\frac{{{T}_{L}}}{{{T}_{H}}}\] ?? (i) And \[\frac{1}{3}=1-\frac{{{T}_{L}}-50}{{{T}_{H}}}\] ?.. (ii) From Eq. (i), \[\frac{{{T}_{L}}}{{{T}_{H}}}=\frac{4}{5}\] \[\Rightarrow \] \[{{T}_{H}}=\frac{5}{4}{{T}_{L}}\] Substituting value of\[{{T}_{H}}\]in Eq. (ii), we get \[\frac{1}{3}=1-\frac{{{T}_{L}}-50}{\frac{5}{4}{{T}_{L}}}\] Or \[\frac{4({{T}_{L}}-50)}{5{{T}_{L}}}=\frac{2}{3}\] Or \[{{T}_{L}}-50=\frac{2}{3}\times \frac{5}{4}{{T}_{L}}\] Or \[{{T}_{L}}-\frac{5}{6}{{T}_{L}}=50\] \[\therefore \] \[{{T}_{L}}=50\times 6=300\,K\]
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