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

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

  • question_answer
    An ideal gas heat engine operates in Carnot cycle between 227°C and 127°C. It absorbs \[6\times {{10}^{4}}\] cals of heat at higher temperature. Amount of heat converted to work is  [CBSE PMT 2005]

    A)            \[2.4\times {{10}^{4}}\]cal     

    B)            \[6\times {{10}^{4}}\] cal

    C)            \[1.2\times {{10}^{4}}\] cal    

    D)            \[4.8\times {{10}^{4}}\] cal

    Correct Answer: C

    Solution :

                       \[\eta =\frac{{{T}_{1}}-{{T}_{2}}}{{{T}_{1}}}=\frac{W}{Q}\]Þ \[W=\frac{Q({{T}_{1}}-{{T}_{2}})}{{{T}_{1}}}\]                    \[=\frac{6\times {{10}^{4}}\left[ (227+273)-(273+127) \right]}{(227+273)}\]            \[=\frac{6\times {{10}^{4}}\times 100}{500}\]\[=1.2\times {{10}^{4}}cal\]


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