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Question: At what temperature will aluminium have a resistivity that is three times the resistivity copper has...

At what temperature will aluminium have a resistivity that is three times the resistivity copper has at room temperature?

Explanation

Solution

We are asked to find the temperature at which the temperature of aluminum at which the resistivity becomes three times that of copper at room temperature. We can start by coming up with a proper equation of the above relation. Once we have come up with it, we can easily solve the question by substituting the values of each of the terms thus getting the required solution.
The formula to find the resistivity at a temperature is given by the formula ρ=ρ0(1+αΔT)\rho = {\rho _0}\left( {1 + \alpha \Delta T} \right)
Where ΔT\Delta T is the change in temperature
ρ0{\rho _0} is the initial resistivity of aluminium.

Complete answer:
Let us start by writing down an equation for the relation given in the question.
ρAl=3ρCu{\rho _{Al}} = 3{\rho _{Cu}}
Where ρAl{\rho _{Al}} and ρCu{\rho _{Cu}} are the resistivity of aluminium and resistivity of copper respectively.
Now we can use the formula ρ=ρ0(1+αΔT)\rho = {\rho _0}\left( {1 + \alpha \Delta T} \right) and substitute the value to get
ρ0Al(1+αΔT)=3ρCu{\rho _{{0_{Al}}}}\left( {1 + \alpha \Delta T} \right) = 3{\rho _{Cu}}
Where ΔT\Delta T is the change in temperature
ρ0Al{\rho _{{0_{Al}}}} is the initial resistivity of aluminium
Bringing the terms to the other side we get αΔT=3ρCu−ρ0Alρ0Al\alpha \Delta T = \dfrac{{3{\rho _{Cu}} - {\rho _{{0_{Al}}}}}}{{{\rho _{{0_{Al}}}}}}
Now we bring alpha to the other side and get the equation for the change in temperature and it will be as ΔT=3ρCu−ρ0Alαρ0Al\Delta T = \dfrac{{3{\rho _{Cu}} - {\rho _{{0_{Al}}}}}}{{\alpha {\rho _{{0_{Al}}}}}}
Now that we have found the equation for the change in temperature, we can use it or substitute it in the formula T=ΔT+T0T = \Delta T + {T_0} and get the temperature as T=ΔT+T0=3ρCu−ρ0Alαρ0Al+T0T = \Delta T + {T_0} = \dfrac{{3{\rho _{Cu}} - {\rho _{{0_{Al}}}}}}{{\alpha {\rho _{{0_{Al}}}}}} + {T_0}
We can now substitute the values and get the temperature as T=ΔT+T0=3ρCu−ρ0Alαρ0Al+T0⇒3×1.7×10−8−2.82×10−82.82×10−8=227∘CT = \Delta T + {T_0} = \dfrac{{3{\rho _{Cu}} - {\rho _{{0_{Al}}}}}}{{\alpha {\rho _{{0_{Al}}}}}} + {T_0} \Rightarrow \dfrac{{3 \times 1.7 \times {{10}^{ - 8}} - 2.82 \times {{10}^{ - 8}}}}{{2.82 \times {{10}^{ - 8}}}} = 227^\circ C .

Note:
Resistivity is defined as the electrical resistance of a material of unit cross-sectional area and unit length. It is a distinctive property of every material; resistivity is useful in comparing materials on the basis of their ability to conduct electric currents. High resistivity means the materials are poor conductors.