Question

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?

Answer 1

Given the operating voltage V and power consumed P, the resistance of the immersion heater,
$R=\frac{V^2}{P}=\frac{{\left(220\right)}^2}{1000}=48.4\mathrm{\Omega }$
Mass of water, m = $\frac{1}{100}$ × 1000 = 10 Kg
Specific heat of water, s = 4200 Jkg^$-$1 K^$-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,
$\left(\frac{V^2}{R}\right)$ × t × 60% = 1050000 J
$\Rightarrow \frac{(220{)}^2}{48.4}\times t\times \frac{60}{100}=1050000$
⇒ t = 29.17 minutes