Question

An electric bulb is rated 200 W – 200 V. It is immersed in 200 g of oil (SHC 0.8 Jg⁻¹°C⁻¹) at 10°C. The bulb is switched on for 2 minutes. If all the electric energy is absorbed in the form of heat energy by the oil, calculate:

1. Current flowing through the bulb.
2. Resistance of the filament of the bulb.
3. Final temperature of the oil.

Answer 1

m = 200 g; P = 200 W; V = 200 V; t = 2 min = 120 s; $$c = 0.8\frac{\text{J}}{\text{g}^\circ\text{C}}$$

Let the rise in temperature of the oil be θ_R.

(a) P = I × V

\[\therefore \text{I}  =\frac{\text{P}}{\text{~V}}\]

\[=\frac{200}{200}\]

= 1 A

\[\text{(b) }\text{R} = \frac{\text{V}}{\text{I}}\]

\[= \frac{200}{1}\]

= 200 Ω

(c) Heat energy supplied by the bulb = P × t

= 200 × 120

= 24000 J

Heat energy absorbed by the oil = m c θ_R

= 200 × 0.8 × θ_R

∴ By the principle of calorimetry

Heat absorbed by the oil = Heat supplied by the bulb

200 × 0.8 × θ_R = 24000

\[\therefore \theta_{R}=\frac{24000}{200 \times 0.8}\]

= 150°C

∴ Final temperature = initial temperature + rise in temperature (θ_R)

= 10°C + 150°C

= 160°С