Question

An electric heater having a heating coil of 150 ohms connected with a supply voltage of 220 V is used to heat water. The time taken to increase the temperature of 1 kg of water by 20ºC to 60ºC is

Answer 1

Step 1: Given data

1. The resistance of the coil is, $R=150\Omega$
2. The mass of the water is, $m=1kg$
3. Supply voltage is $V=220volt$
4. The increase in temperature is $∆T=\left(60-20\right)ºC=40ºC$.

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,

$$\begin{gathered} H=I^{2}RT=\frac{V^2}{R}\times T(V=IR) \\ orH=\frac{{220}^2}{150}\times T=\frac{48400}{150}=\frac{4840}{15}T \\ orH==\frac{4840}{15}T..........\left(1\right) \end{gathered}$$

Step 4: Calculation of heat loss

Again, heat loss,

$$\begin{gathered} ∆H=ms∆T=1\times 4200\times \left(60-20\right) \\ orH=4200\times 40 \\ orH=168000.................\left(2\right) \end{gathered}$$ (specific heat of water is $s=4200J/kg°C$ )

Step 5: Calculation of time

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

$$\begin{gathered} \frac{4840}{15}\times T=168000 \\ orT=\left(\frac{168000}{4840}\times 15\right)=520.67 \\ orT=520.67seconds. \end{gathered}$$

Therefore, the time taken to increase the temperature of 1 kg of water from 20ºC to 60ºC is $520.67seconds.$