VMMC VMMC Medical Solved Paper-2007

  • question_answer
    The temperature of sink of a Carnot engine is 27°C. If the efficiency of engine be 25%, then the temperature of source must be

    A) \[27{}^\circ C\] 

    B)                        \[127{}^\circ C\]

    C) \[227{}^\circ C\]                              

    D)        \[327{}^\circ C\]

    Correct Answer: B

    Solution :

    Efficiency of Car not engine is given by, \[\eta =1-\frac{{{T}_{2}}}{{{T}_{1}}}\] where \[{{\text{T}}_{\text{2}}}\] is temperature of sink in kelvin and \[{{\text{T}}_{1}}\]is temperature of source in kelvin. Given,     \[\eta =25%=\frac{25}{100}=0.25\]                                          \[{{T}_{2}}=27{{\,}^{o}}C=27+273=300\,K\]                 \[\therefore \]  \[0.25=1-\frac{300}{{{T}_{1}}}\]                 \[\Rightarrow \]                               \[\frac{300}{{{T}_{1}}}=1-0.25\] \[\Rightarrow \]                               \[\frac{300}{{{T}_{1}}}=0.75\] \[\Rightarrow \]               \[{{T}_{1}}=\frac{300}{0.75}=400\,K=127\,{{\,}^{o}}C\]


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