A) 200 K, 250 K
B) 250 K, 200 K
C) 300 K, 250 K
D) 300 K, 200 K
Correct Answer: B
Solution :
By Carnot ideal heat engine \[\eta =1\,-\frac{{{T}_{2}}}{{{T}_{1}}}\] Where \[{{\eta }_{1}}=0.2,\,{{\eta }_{2}}=0.4\] For the first condition, \[{{\eta }_{1}}\,=1-\frac{{{T}_{2}}}{{{T}_{1}}}\Rightarrow \,0.2\,=1-\frac{{{T}_{2}}}{{{T}_{1}}}\]or \[0.2=\frac{{{T}_{1}}-{{T}_{2}}}{{{T}_{1}}}\]for the second condition \[{{\eta }_{2}}\,=1-\frac{{{T}_{2}}-50}{{{T}_{1}}}\Rightarrow \,0.4=\,\frac{{{T}_{1}}-({{T}_{2}}-50)}{{{T}_{1}}}\] \[0.4=\frac{{{T}_{1}}-{{T}_{2}}+50}{{{T}_{1}}}\,\Rightarrow \,0.4=\frac{{{T}_{1}}-{{T}_{2}}}{{{T}_{1}}}\,+\frac{50}{{{T}_{1}}}\] \[0.4\,=0.2+\frac{50}{{{T}_{1}}}\] \[0.4\,-0.2=\frac{50}{{{T}_{1}}}\] \[{{T}_{1}}=\frac{50}{2}\Rightarrow \,{{T}_{1}}=250\,K\] Putting the value of \[{{T}_{1}}\] in eq (i) we get \[0.2=\frac{{{T}_{1}}-{{T}_{2}}}{{{T}_{1}}}\Rightarrow \,{{T}_{2}}-{{T}_{1}}\,-0.2{{T}_{1}}\] \[{{T}_{2}}=250-0.2\times 250\] \[{{T}_{2}}=250\,-50\Rightarrow \,{{T}_{2}}\,=200\,K\] \[{{T}_{1}}\,=250K,\,\,{{T}_{2}}=200\,K\]
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