Question

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

(Given Heat capacity of water$=4186\mathrm{J}/\mathrm{kg}°\mathrm{C}$)

Answer 1

Step 1: Write the given data

Potential difference $\mathrm{V}=220\mathrm{V}$

Resistance of the coil $\mathrm{R}=150\mathrm{\Omega }$

Mass$=1\mathrm{kg}$

Specific heat capacity of water$=4186\mathrm{J}/\mathrm{kg}°\mathrm{C}$

Step 2: Determine the change in temperature

Change in temperature,$∆\mathrm{T}={\mathrm{T}}_2-{\mathrm{T}}_1$

$$\begin{gathered} ∆\mathrm{T}=60°\mathrm{C}-20°\mathrm{C} \\ ∆\mathrm{T}=40°\mathrm{C} \end{gathered}$$

Therefore, the change in temperature is $40°\mathrm{C}$.

Step 3: Determine the heat generated by heater

Energy(heat) generated, $H=\frac{{\mathrm{V}}^2\mathrm{t}}{\mathrm{R}}$

$H=\frac{{\left(220\mathrm{V}\right)}^{2}t}{150\mathrm{\Omega }}$

$H=\frac{4840t}{15}$ …… (1)

So, heat generated is $\frac{\left(4840\mathrm{V}\right)t}{\left(15\mathrm{\Omega }\right)}$

Step 4: Determine the total heat transfer

$\mathrm{Heat}\mathrm{transfer}=\mathrm{mass}\times \mathrm{specific}\mathrm{heat}\mathrm{capacity}\times \mathrm{temperature}\mathrm{difference}$

Substitute the values and solve.

$$\begin{gathered} H=1\times 4186\times 40 \\ H=167440\mathrm{Joules} \end{gathered}$$

Step 5: Determine the time taken to heat 1 kg of water

Heat generated = Heat transfer

$$\begin{gathered} 167440=\frac{4840t}{15} \\ t=\frac{\left(167440\right)15}{4840} \\ t=518.92\sec \end{gathered}$$

Therefore time takes to heat up the coil is $518.92\mathrm{seconds}$.