Answer :
Given that the potential difference is V = 1.5 V.
The length of the wire is l = 1.5 m.
The cross-sectional area is
[tex]\begin{gathered} A\text{ = 0.6 }mm^2 \\ =0.6\times10^{-6}m^2 \end{gathered}[/tex]The resistivity of the wire is
[tex]\rho\text{ = 5.25}\times10^{-8}\Omega\text{ m }[/tex]We have to find the power dissipated in the wire.
First, we need to calculate resistance.
The resistance can be calculated as
[tex]\begin{gathered} R\text{ = }\rho\frac{l}{A} \\ =5.25\times10^{-8}\times\frac{1.5}{0.6\times10^{-6}} \\ =0.13125\Omega \end{gathered}[/tex]The formula to calculate power is
[tex]P\text{ =}\frac{V^2}{R}[/tex]Substituting the values, the power will be
[tex]\begin{gathered} P\text{ = }\frac{(1.5)^2}{0.13125} \\ =17.1\text{ W} \end{gathered}[/tex]Thus, the power dissipated in the wire is 17.1 W