Question

An immersion rod having resistance of 50 Ω is connected to 220V main supply. Assuming that all the energy generated goes to heat the water, calculate the time taken to heat 5 kg water from 30°C to 100°C.

Answer 1

We know that  P = VI

∴ I = `"P"/"V" = 220/50 = 4.4`

Heat produced in an immersion heater

H = I²Rt

H = (4.4)² × 50 × t

H = 968 t Joules

and Heat needed = mcθ

where C is the sp. heat capacity of water

= 4.2 × 10³ J/kgC = 5 × 4.2 × 10³ × (100 - 30)

= 5 × 4.2 × 10³ × 70 = 1470000 J.

But Heat lost = Heat gained

968 t = 1470000

t = `1470000/968`

t = 1518.59 secs

t = 1518.6 secs

t = 25.326 min.

t = 25.33 mins.