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Question: What is the effect of temperature on the resistance of a metal? The resistance of a platinum resista...

What is the effect of temperature on the resistance of a metal? The resistance of a platinum resistance thermometer at 0C0{}^\circ C temperature is 3.0Ω3.0\Omega and at 100C100{}^\circ C it is 3.75Ω3.75\Omega . Its resistance at an unknown temperature is 3.15Ω3.15\Omega . Find the value of the unknown temperature.

Explanation

Solution

We know that the resistance changes with increase in temperature but different materials show different trends. So, we could find the trend shown by the metals. Now recall the expression for temperature dependence of resistance and then by substituting accordingly find the value of unknown temperature using given values.

Formula used:
Temperature dependence of resistance,
R=R0(1+αΔT)R={{R}_{0}}\left( 1+\alpha \Delta T \right)

Complete answer:
In the question, we are asked to find the effect of temperature on the resistance of a metal. Also, we are given the resistance at 0C0{}^\circ C and 100C100{}^\circ C as 3.0Ω3.0\Omega and 3.75Ω3.75\Omega respectively. Then, we are asked to find the value of the temperature at which the resistance is 3.15Ω3.15\Omega .
When the temperature of the metal is increased, the ions in the metal start oscillating about their mean positions by acquiring energy. These ions collide with the electrons and thus affect the electron flow. So we could say that the resistance of metals increases with increase in temperature.
We know that the temperature dependence of a material is given by,
R=R0(1+αΔT)R={{R}_{0}}\left( 1+\alpha \Delta T \right) ……………………………………. (1)
Let, R0{{R}_{0}} be the resistance at 0C0{}^\circ C and R1{{R}_{1}} be the resistance at100C100{}^\circ C, then,
R0=3.0Ω{{R}_{0}}=3.0\Omega
R1=3.75Ω{{R}_{1}}=3.75\Omega
So, from (1) we have,
3.75=3.0(1+α(1000))3.75=3.0\left( 1+\alpha \left( 100-0 \right) \right)
100α+1=1.25\Rightarrow 100\alpha +1=1.25
α=0.25×102C1\Rightarrow \alpha =0.25\times {{10}^{-2}}{}^\circ {{C}^{-1}} ……………………………………… (2)
Now let R2=3.15Ω{{R}_{2}}=3.15\Omega be the resistance at the unknown temperature T then, by (1) we have,
3.15=3.0(1+(0.25×102)(T0))3.15=3.0\left( 1+\left( 0.25\times {{10}^{-2}} \right)\left( T-0 \right) \right)
1+0.0025T=1.05\Rightarrow 1+0.0025T=1.05
T=20C\therefore T=20{}^\circ C

Hence, we found the value of temperature at which the resistance is 3.15Ω3.15\Omega to be 20C20{}^\circ C.

Note:
Not every material shows an increase in the value of resistance with the increase in temperature. Every metal and almost all alloys show this trend of increase in resistance with increase of temperature. But for the semiconductors, electrolytes and insulators, the resistance decreases with increase in temperature.