Question

30 people gather in a $10\text{m}\times 5\text{m}\times 3\text{m}$ room for a confidential meeting. The room is completely sealed off and insulated. Calculate the rise in temperature of the room in half an hour. Assume that the average energy thrown off by the body of a person is $2500\text{kcal/day}$, density of air is $1.2{\text{kg/m}}^3$ and specific heat capacity of air at constant volume is $0.24\text{kcal kg}{{}^{-1}}^{\circ }{\text{C}}^{-1}$.
[Neglect volume occupied by the people].

• A. ${18}^{\circ }\text{C}$
• B. ${36}^{\circ }\text{C}$
• C. ${54}^{\circ }\text{C}$
• D. ${68}^{\circ }\text{C}$

Answer 1

The correct option is B ${36}^{\circ }\text{C}$

Given that, the room is closed and insulated i.e there is no change in volume.
Thus, the work done by air in the room $W=0$
$\therefore$ From first law of thermodynamics, we can say that, $\Delta U=\Delta Q$
From the data given in the question,
Total heat released into the room by $30$ people
$Q=30\times 2500=75000\text{kcal/day}$
$=\frac{75000}{24\times 60}\text{kcal/min}=52\text{kcal/min}$
Volume of air in the room $V=10\times 5\times 3=150{\text{m}}^3$
$\therefore$ Mass of air in the room $m=1.20\times 150=180\text{kg}$

We know that $\Delta Q=mc\Delta T$
In 30 minutes,
$mc\Delta T=52\times 30\text{kcal}$
Thus, rise in temperature of the room in $30\text{min}$
$\Delta T=\frac{52\times 30}{180\times 0.24}={36}^{\circ }\text{C}$
Thus, option (b) is the correct answer.