2. Sample Papers
  3. KVPY

KVPY Sample Paper KVPY Stream-SX Model Paper-2

  • question_answer
    Consider a class room of dimension \[5\times 10\times 3\,{{m}^{3}}\] at temperature \[20{}^\circ C\] and pressure 1 atm. There are 50 peoples in the room, each losing energy at the average of 150 e=watt. Assuming that the walls, ceiling, floor and furniture perfectly insulated and none of them absorbing heat: the time needed for rising the temperature of air in the room to body temperature,\[i.e.,37{}^\circ C\] will be (for air Cp=7/2 R. loss of air to the outside as the temperature rises may be neglected)

    A) 420.25             

    B) 415.55

    C) 411.35 

    D) 408.35

    Correct Answer: C

    Solution :

    [c]
    Volume of air in the room \[=5\times 10\times 3=150{{m}^{3}}=150\times {{10}^{6}}c{{m}^{3}}\]
    \[\therefore \] mole of air =\[n=\frac{PV}{RT}\]\[=\frac{1\times 150\times {{10}^{6}}}{{{10}^{3}}\times 0.0821\times 293}=6.236\times {{10}^{3}}\] 
    Also, \[\frac{\delta H}{\delta T}={{C}_{P}}=\frac{\Delta H}{\Delta T}\]   (for mole)
    \[\therefore \Delta H={{n}^{.}}{{C}_{P}}^{.}\Delta T\]                        (for mole)\[=6.236\times {{10}^{3}}\times \frac{7}{2}\times 8.314\times \left( 310-293 \right)\]\[=3.085\times {{10}^{6}}J\]
    Thus, heat needed to heat the room to \[37{}^\circ ~\]is \[3.085\times {{10}^{6}}J\]
    Also, heat released by 50people
    \[\begin{align}   & =15\times 150J/\sec =7500J/\sec  \\  & \left( 1watt=1J/\sec  \right) \\ \end{align}\]
    7500 J heat is provided in 1 sec
    \[\therefore 3.085\times {{10}^{6}}J\]Heat will be provided in \[\frac{1\times 3.085\times {{10}^{6}}}{7500}=411.3sec\]           


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