2. Questions Bank
  3. JEE Main & Advanced
  4. Physics
  5. Thermodynamical Processes

JEE Main & Advanced Physics Thermodynamical Processes Question Bank Topic Test - Thermodynamical Processes

  • question_answer
    Two carnot’s engines A and B are operated in series. The first one A receives heat at 1200 K and rejects to a reservoir at T and K. The second engine B receives the heat rejected by the first engine and in turn rejects to a heat reservoir at 300 K. Calculate the value of T, when work outputs of the two engines are equal.

    A) 600 K

    B) 750 K

    C) 450 K

    D) 900 K

    Correct Answer: B

    Solution :

    [b]\[\frac{W}{{{Q}_{1}}+W}={{\eta }_{1}}\] \[\frac{W}{{{Q}_{1}}}={{\eta }_{2}}\] 
    \[\frac{W}{{{Q}_{1}}}=1-\frac{300}{T}\]
    \[\Rightarrow \] \[\frac{1}{1+\frac{T}{T-300}}=\frac{1200-T}{1200}\]
    \[\Rightarrow \]\[(T-300)1200=(1200-T)(2T-300)\]
    \[T=750K\]


You need to login to perform this action.
You will be redirected in 3 sec spinner