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## Q. 2.21

30 people attend a party in a small room of size 5m × 10m × 3m. Each person gives about 400kJ of heat per hour. Assuming the room to be completely sealed and insulated, calculate the air temperature rise within 15 minutes. Assume for air, $C_ν$= 0.718 kJ/kg. K and R = 0.287 kJ/kg. K. Each person occupies a volume of 0.07 m³ and initial room conditions are 1 bar at 20°C.

## Verified Solution

Volume of room, $V_r=5 \times 10 \times 3=150 m ^3$

Volume of air, $V_a=150-30 \times 0.07=147.9 m ^3$

Mass of air, $m_a=\frac{p V_a}{R T}=\frac{1 \times 10^2 \times 147.9}{0.287 \times 293}=175.88 \ kg$

From first law of thermodynamics,

Q – W = dU

Assuming constant volume process, W = 0

$\therefore$           Q = dU = 400 × 30 =12000 kJ/h

For 15 minutes period,

dU = 12000 × 15/60 = 3000 kJ

\begin{aligned}d U &=m_a c_v d T \\d T &=\frac{3000}{175.88 \times 0.718}=23.76^{\circ} C\end{aligned}