Question

Consider a classroom of dimensions $(5\times 10\times 3)$m${}^3$ at temperature ${20}^o$C and pressure $1$ atm. There are $50$ people in the room, each losing energy at the average of $150$ watts. Assuming that the walls, ceiling, floor, and furniture are perfectly insulated and none of them is absorbing heat. How much time will be needed for raising the temperature of air in the room to the body temperature (37${}^{o}C$)? [For air $C_p=\frac{7}{2}R$ and neglect the loss of air to the outside as the temperature rises]

• A. $422$ sec
• B. $411.3$ sec
• C. $421.1$ sec
• D. $413.1$ sec

Answer 1

The correct option is B $411.3$ sec

The volume of air present in the room is $5\times 10\times 3=150m^3=150\times {10}^3$ $litre$ ($\because 1m^3=1000L$)

The number of moles of air is calculated using the ideal gas equation.

$n=\frac{PV}{RT}=\frac{1\times 150\times {10}^3}{0.0821\times 293}=6.236\times {10}^3$

For 1 mole of air, ${\left(\frac{\partial H}{\partial T}\right)}_P=C_p=\frac{\Delta H}{\Delta T}$

For n moles of air, $\therefore \Delta H=n\times C_p\times \Delta T$

Substitute values in the above expression,
$\Delta H=6.236\times {10}^3\times \frac{7}{2}\times 8.314\times (310-293)=3.085\times {10}^{6}J$

Thus, the heat need to increase the temperature of the room to ${37}^{\circ }$C is $=3.085\times {10}^{6}J$

The heat released by $50$ people is $=150\times 50J/sec=7500J/sec$. This is the heat provided in 1 second.

$\therefore$ The time required to provide $3.085\times {10}^{6}J$ heat is $\frac{1\times 3.085\times {10}^6}{7500}=411.3$ sec.

Hence, the correct option is $\text{B}$