A) \[133{}^\circ C\]
B) \[143{}^\circ C\]
C) \[100{}^\circ C\]
D) \[373{}^\circ C\]
Correct Answer: D
Solution :
\[\eta =\frac{Net\text{ }work\text{ }done\text{ }per\text{ }cycle}{Total\text{ }amount\text{ }of\text{ }heat\text{ }absorbed\text{ }per\text{ }cycle}\] or \[\eta =\frac{W}{{{Q}_{1}}}\] \[\therefore \] \[\eta =\frac{{{Q}_{1}}-{{Q}_{2}}}{{{Q}_{1}}}\] Or \[\eta =1-\frac{{{Q}_{2}}}{{{Q}_{1}}}\] as \[\frac{{{Q}_{2}}}{{{Q}_{1}}}=\frac{{{T}_{2}}}{{{T}_{1}}}\] \[\therefore \] \[\eta =1-\frac{{{T}_{2}}}{{{T}_{1}}}\] ?.(i) where,\[{{T}_{2}}\]is temperature of sink and\[{{T}_{1}}\]is the temperature of the sources. Here, \[\eta =50%\] \[{{T}_{2}}=50{}^\circ C=273+50\] \[=323K\] Putting these values in Eq. (i), we get \[\frac{50}{100}=1-\frac{323}{{{T}_{1}}}\] \[T=646K=373{}^\circ C\]
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