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 \[\frac{1\times 3.085\times {{10}^{6}}}{7500}=411.3sec\] |
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