A bulb of 100 W capacity was switched on in a thermally insulated room of size (2.5 × 3 × 3) m^{3} and temperature 20°C. Find the temperature of the room at the end of 24 hours.

This question was previously asked in

AAI JE (Technical) Official Paper 2018

Option 3 : 470 °C

ST 2: Strength of materials

2026

15 Questions
15 Marks
15 Mins

__Concept__:

According to First Law of Thermodynamics:

δQ = dU + δW

For insulated room δQ = 0

∴ dU + δW = 0

∴ dU = -δW [-ve implies that work is done on the system]

Change in internal energy (dU) = mcvdT

where m = mass, cv = specific heat capacity at constant volume, dT = difference between final and initial temperature.

__Calculation__:

__Given__:

Dimension of room = 2.5 m × 3 m × 3 m, Initial temperature Ti = 20°C, Time = 24 hours ⇒ 24 × 3600 sec, ρ = 1.2 kg/m3, cv = 0.717 kJ/kg, Electric bulb heating rate = 100 W.

The work done on the room by electric heater is δW

δW = 100 × 24 × 3600 = 8.64 × 106 J

Volume of room = 2.5 × 3 × 3 = 22.5 m3

Mass of air = ρV ⇒ 1.2 × 22.5 = 27 kg.

Let T2 be the final temperature of the gas.

Because the volume of the room is constant

δW = dU

δW = mCv(T2 – T1)

8.64 × 106 = 27 × 0.717 × 103 × (T2 – 20)

T2 = 466.30° C ≈ 470°C