Question

An electric kettle takes $4A$ current at 220V. How much time will it take to boil $1kg$ of water from room temperature $20{}^{o}C$? (The temperature of boiling water is $100{}^{o}C$).

Answer 1

Step 1: Given data

1. The mass of the water is, $m=1kg=1000g.$
2. Current through the kettle is, $I=4A.$
3. Supply voltage is $V=220volt$
4. The increase in temperature is $∆T=\left(100-20\right)ºC=80ºC$.

Step 2: Joule's law of heating and heat loss

1. Joule's law states that when a current I is flowing in a conductor of resistance R for a time t sec, then the produced heat on the conductor is directly proportional to the square of the electric current.
2. Joule's law is defined by the form, $H=I^{2}Rt$ , where, Q is the produced heat on the conductor.
3. Heat loss by a conductor is defined by the form, $∆H=ms∆T$, where, $m$ is the mass of the conductor, $s$ is the specific heat, and $∆T$ is the temperature difference.

Step 3: Calculation of produced heat

Now applying Joule's law, heat generated by the conductor,

$$\begin{gathered} H=I^{2}Rt=V\times I\times t(\sin ce,V=IR) \\ orH=\left(220\right)\times 4\times TCal(1Cal=4.2joule) \\ orH=\frac{880T}{4.2}Joule...........\left(1\right) \end{gathered}$$

Step 4: Calculation of heat loss

Again, heat loss,

$$\begin{gathered} ∆H=ms∆T=1000\times 1\times \left(100-20\right) \\ orH=1000\times 80 \\ orH=80000.................\left(2\right) \end{gathered}$$ (specific heat of water is $s=1C/g°C$ )

Step 5: Calculation of time

Using the property of heat and calorimetry
Heat loss = Heat gain. So, from equations 1 and 2

$$\begin{gathered} \frac{880\times T}{4.2}=80000 \\ orT=\frac{80000}{880}\times 4.2=381 \\ orT=381seconds. \\ orT=\frac{381}{60}=6.36min \\ orT=6.36min. \end{gathered}$$

Therefore, the time taken to increase the temperature of 1 kg of water from 20ºC to 100ºC is $6.36min$.