Question

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

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

Answer 1

Concept Used: Heat

Step 1: Given Data,

Potential Difference $=220\mathrm{V}$

Resistance R$=110\mathrm{\Omega }$

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

Mass =$1\mathrm{kg}$

Change in temperature=$T_2-T_1$

Change in temperature= $70°\mathrm{C}-20°\mathrm{C}=50°\mathrm{C}$

Step 2: Calculate the time,

$$\begin{gathered} \mathrm{Heat}\mathrm{loss}=\mathrm{Mass}\times \mathrm{Specific}\mathrm{heat}\mathrm{capacity}\times \mathrm{change}\mathrm{in}\mathrm{temperature} \\ =1\mathrm{kg}\times 4186\mathrm{J}/\mathrm{kg}°\mathrm{C}\times 50°\mathrm{C} \end{gathered}$$

$$\begin{gathered} \mathrm{Heat}\mathrm{Produced}=\frac{{\mathrm{V}}^2\mathrm{t}}{\mathrm{R}} \\ =\frac{{\left(220\right)}^2\times \mathrm{t}}{110} \end{gathered}$$

To find time, $\mathrm{Heat}\mathrm{produced}=\mathrm{Heat}\mathrm{loss}$

$$\begin{gathered} 440\mathrm{t}=4186\times 50 \\ \mathrm{t}=475.7\mathrm{s} \end{gathered}$$

Time taken to het up the coil is $475.7\mathrm{s}$