Question

A \(100~\text{W}\) electric bulb was switched on in a \(2.5~\text{m}\times 3~\text{m}\times3~\text{m}\) size thermally insulated room having a temperature of \(20~^{\circ}\text{C}\). The room temperature at the end of \(24~\text{hours}\) will be (Assume, air to be ideal in nature)

[Data required: specific heat of air at constant volume \(c_v=0.717~\text{kJ/kgK}\) and density of air, \(\rho =1.2~\text{kg/m}^3\)]

Answer 1

During \(24~\text{hours}\) heat liberated by bulb is
\(Q = 100 × 24 × 3600\)

\(\therefore Q= 8.64 × 10^6~\text{J}\)

Given, density of air,
\(\rho =1.2~\text{kg/m}^3\)

Volume of room,
\(V = 2.5×3×3=22.5~\text{m}^3\)

Mass of air,
\(m=1.2×22.5=27~\text{kg}\)

Let, \(T_2\) be the final temperature of gas,

Also, Because volume of room is constant,

\(Q=mc_v(T_2-T_1)\)

where, \(c_v=\)specific heat of air at constant volume\(=0.717~\text{kJ/kgK}\)

Substituting the values, in the above equation,

\(\Rightarrow 8.64×10^6=27×0.717×1000(T_2-20)\)

\(\therefore T_2 = 470~^{\circ}\text{C}\)

Hence, option (d) is correct answer.