A 100 W electric bulb was switched on in a 2.5 m × 3 m × 3 m size thermally insulated room having a temperature of 20°C. The room temperature at the end of 24 hours will be. (Take ​density of air = 1.2 kg/m3 and cv of air = 0.717 kJ/kg.

CONCEPT:According to First Law of Thermodynamics:δQ = dU + δWFor insulated ROOM δQ = 0∴ dU + δW = 0∴ dU = -δW [-ve implies that work is done on the system]Change in internal energy (dU) = mcvdTwhere 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 δWδW = 100 × 24 × 3600 = 8.64 × 106 JVolume of room = 2.5 × 3 × 3 = 22.5 m3Mass 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 constantδW = dUδW = mCv(T2 – T1)8.64 × 106 = 27 × 0.717 × 103 × (T2 – 20)T2 = 466.30° C ≈ 470°C