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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.
Solution
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.
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