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JEE Main & Advanced Physics Thermodynamical Processes Question Bank Heat Engine Refrigerator and Second Law of Thermodynamics

  • question_answer
    A Carnot engine absorbs an amount Q of heat from a reservoir at an abosolute temperature T and rejects heat to a sink at a temperature of T/3. The amount of heat rejected is [UPSEAT 2004]

    A)            Q / 4                                        

    B)            Q / 3

    C)            Q / 2                                        

    D)            2Q / 3

    Correct Answer: B

    Solution :

                       \[\because \ \eta =1-\frac{{{T}_{2}}}{{{T}_{1}}}=\frac{W}{{{Q}_{1}}}\]\[{{T}_{2}}=300\,K\] where \[{{Q}_{1}}=\] heat absorbed, \[{{Q}_{2}}=\] heat rejected Þ \[1-\frac{T/3}{T}=\frac{W}{{{Q}_{1}}}\] Þ \[\frac{2}{3}=\frac{W}{{{Q}_{1}}}=\frac{{{Q}_{1}}-{{Q}_{2}}}{{{Q}_{1}}}\] Þ \[\frac{2}{3}=1-\frac{{{Q}_{2}}}{{{Q}_{1}}}\] Þ \[\frac{{{Q}_{2}}}{{{Q}_{1}}}=\frac{1}{3}\] Þ \[{{Q}_{2}}=\frac{{{Q}_{1}}}{3}=\frac{Q}{3}\]


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