Question

A metallic coil connected to a $220\text{V}$ supply, has a resistance of $110\Omega$ (while hot). How long will this coil take to heat $1\text{kg}$ of water from $20$ to ${70}^{\circ }C$? Assume that whole of heat produced by the coil is taken up by water.

(Heat capacity Of Water $=4186{\text{J/kg}}^{\circ }\text{C}$)

Answer 1

Step 1: Coil,

The coil is the wire which is wound into a series of loops. The coil produces heat when the current is passed through it, and this heat can be used for different purposes like boiling water.

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

Resistor,$R=110\Omega$

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

Change in temperature

,$$\begin{gathered} \Delta T=70-20 \\ \Delta T={50}^{\circ }C \end{gathered}$$

Step 2:

The energy consumed by the heater is given as,

$H=\frac{V^2}{R}\times t$ ……(1)

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

Energy absorbed by water is given by

$H=ms\Delta T$ …..(2)

Step 3:

All heat is taken by water, from the equation (1) and (2)

$\frac{V^2}{R}\times t=ms\Delta T$

Substitute the values and solve for time $t$,

$$\begin{gathered} \frac{{220}^2}{110}\times t=1\times 4186\times 50 \\ t=\frac{209300\times 110}{{220}^2} \\ t=4.324\times 110 \\ t=475.68\text{sec} \end{gathered}$$

Hence, the time taken by the coil to heat $1\text{kg}$ of water from ${20}^{\circ }C$ to ${70}^{\circ }C$ is $475.68\text{seconds}$.