A) \[133{}^\circ C\]
B) \[143{}^\circ C\]
C) \[100{}^\circ C\]
D) \[373{}^\circ C\]
Correct Answer: D
Solution :
Thermal efficiency of a Carnots engine is defined as the ratio of net work done per cycle by the engine to the total amount of heat absorbed per cycle by the working substance from the source. It is denoted by \[\eta \] Thus, \[\eta =\frac{Net\text{ }work\text{ }done/cycle}{Total\text{ }amount\text{ }of\text{ }heat\text{ }absorbed/cycle}\]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 source. Here, \[\eta =50%\] \[{{T}_{2}}={{50}^{o}}C=273+50=323K\] Putting these values in Eq. (i), we get \[\frac{50}{100}=1-\frac{323}{{{T}_{1}}}\] or \[\frac{323}{{{T}_{1}}}=1-\frac{1}{2}=\frac{1}{2}\] \[\therefore \] \[{{T}_{1}}=646K={{373}^{o}}C\]
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