Question

A $100\text{W}$ electric bulb was switched on in a $2.5\text{m}\times 3\text{m}\times 3\text{m}$ 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, $\rho =1.2{\text{kg/m}}^3$]

• A. $321{}^{\circ }\text{C}$
• B. $341{}^{\circ }\text{C}$
• C. $450{}^{\circ }\text{C}$
• D. $470{}^{\circ }\text{C}$

Answer 1

The correct option is D $470{}^{\circ }\text{C}$

During $24\text{hours}$ heat liberated by bulb is
$Q=100\times 24\times 3600$

$\therefore Q=8.64\times {10}^6\text{J}$

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

Volume of room,
$V=2.5\times 3\times 3=22.5{\text{m}}^3$

Mass of air,
$m=1.2\times 22.5=27\text{kg}$

Let, $T_2$ be the final temperature of gas,

Also, Because volume of room is constant,

$Q=m{c}_v(T_2-T_1)$

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

Substituting the values, in the above equation,

$\Rightarrow 8.64\times {10}^6=27\times 0.717\times 1000(T_2-20)$

$\therefore T_2=470{}^{\circ }\text{C}$

Hence, option (d) is correct answer.