# An Immersion Heater Rated 1000 W, 220 V is Used to Heat 0.01 M3 Of Water. Assuming that the Power is Supplied at 220 V and 60% of the Power Supplied is Used to Heat the Water - Physics

Sum

An immersion heater rated 1000 W, 220 V is used to heat 0.01 m3 of water. Assuming that the power is supplied at 220 V and 60% of the power supplied is used to heat the water, how long will it take to increase the temperature of the water from 15°C to 40°C?

#### Solution

Given the operating voltage V and power consumed P, the resistance of the immersion heater,

$R = \frac{V^2}{P} = \frac{\left( 220 \right)^2}{1000} = 48 . 4 \Omega$

Mass of water, $m = \frac{1}{100}\times1000=10 kg$

Specific heat of water, s = 4200 Jkg-1K-1

Rise in temperature, θ = 25°C

Heat required to raise the temperature of the given mass of water,

Q = msθ = 10 × 4200 × 25 = 1050000 J

Let t be the time taken to increase the temperature of water. The heat liberated is only 60%. So,

$\left( \frac{V^2}{R} \right)\times t \times 60\% = 1050000 J$

$\Rightarrow \frac{(220 )^2}{48 . 4} \times t \times \frac{60}{100} = 1050000$

⇒ t = 29.17 minutes

Concept: Electrical Energy and Power
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#### APPEARS IN

HC Verma Class 11, Class 12 Concepts of Physics Vol. 2
Chapter 11 Thermal and Chemical Effects of Current
Q 9 | Page 219