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Question: Ten identical cells connected in series are needed to heat a wire of length \(1\,m\) and radius \(r\...

Ten identical cells connected in series are needed to heat a wire of length 1m1\,m and radius rr by 10C{10^ \circ }C in timett . How many cells will be required to heat the wire of length 2m2\,m of the same radius by the same temperature in time tt ?

Explanation

Solution

Here we have to find the number of cells required to heat the wire of given length and to solve this type of question, one must know the concept of specific heat. Here we will put the given data in the formula of heat capacity and then we will compare the obtained relation to get the required solution.

Formula used:
R=ρlAR = \rho \dfrac{l}{A}
Where, rr is the resistance of the wire, ρ\rho is the resistivity of the material, ll is the length of the material and aa is the area of the cross section.

Complete step by step answer:
Let us assume, ρ\rho is the resistivity of the wire, rr is the radius of the wire and EE be the emf of each cell. According to the question, the length of the wire is 1m1\,m.

And 1010 cells each having an emf of EE are connected in series to heat the wire by,
Δt=10C\Delta t = {10^ \circ }C in time tt .
As we know that, wire will take to generate the heat by V2R×t\dfrac{{{V^2}}}{R} \times t
So, we can write, V2R×t=msΔT\dfrac{{{V^2}}}{R} \times t = ms\Delta T
And we know that mass is directly proportional to length. So, we can write,
(10E)2R×t=msΔT\dfrac{{{{\left( {10E} \right)}^2}}}{R} \times t = ms\Delta T -----(1)

And resistance of the same wire of length of 2m2m is,
R=ρ(2)a R=2R R' = \dfrac{{\rho (2)}}{a} \\\ \Rightarrow R' = 2R \\\
Now, let nn be the number of cells having emf EE are connected in series to heat the same wire which is of length 2m2m is,
(nE)2t2R=(2m)sΔT\dfrac{{{{\left( {nE} \right)}^2}t}}{{2R}} = \left( {2m} \right)s\Delta T -----(2)
And now dividing the equation (2) by (1),

\Rightarrow \dfrac{{{n^2}}}{{200}} = 2 \\\ \Rightarrow {n^2} = 400 \\\ \therefore n = 20 \\\ $$ **Hence, $20$ cells will be required to heat the wire of length $2m$ of the same radius by the same temperature in time $t$.** **Note:** Remember the formula of specific heat and when an electric current is conducted through a conductor, heat is generated due to the conductor's impediment to the current flowing inside. The amount of effort required to overcome this impediment to electric current generates heat in that particular conductor.