Question

A heater coil connected to $220\text{V}$ supply, has a resistor of $150\Omega$. How long will this coil take to heat $1\text{kg}$ of water
from $20$ to ${60}^{\circ }C$, assuming that all heat is taken up by water?

Answer 1

Step 1: Heat,

The transfer of energy due to the different temperatures of the bodies is known as heat energy. This energy transfer occurs from a hotter body to a cooler body.

Supply voltage, $V=220\text{V}$

Resistor,$R=150\Omega$

Mass,$m=1\text{kg}$

The heat capacity of water, $s=4186{\text{J/kg}}^{\circ }\text{C}$

Change in temperature,

$$\begin{gathered} \Delta T=60-20 \\ \Delta T={40}^{\circ }C \end{gathered}$$

Step 2:

The generated heat is given by

$H=\frac{V^2}{R}\times t$

Here,$H$ is generated heat and $t$ is the time.

$H=\frac{{220}^2}{150}\times t$

$H=322.666\times t$ ……(1)

Step 3: The heat loss is given by,

$H=ms\Delta T$

substitute the values and solve as

$H=1\times 4186\times 40$

$H=167440\text{J}$ …..(2)

All heat is taken up by the water.

From equations (1) and (2)

$$\begin{gathered} 322.666\times t=167440 \\ t=\frac{167440}{322.666} \\ t=518.92\text{sec} \end{gathered}$$

Hence, the time taken by the coil to heat $1\text{kg}$ of water is $518.9\text{sec}$.