Solution
= Solution
The <Ohmic heating> per unit volume is $\mathbf E\cdot\mathbf J=J^2/\sigma$, uniform in this wire. Multiplying by its volume gives the dissipated <power>
$$
\boxed{P_{\rm loss}=\frac1\sigma\left(\frac I{\pi a^2}\right)^2\pi a^2L
=\frac{I^2L}{\sigma\pi a^2}=I^2R_{\rm wire},}
$$
where the <electrical resistance> is $R_{\rm wire}=L/(\sigma\pi a^2)$. The result is in watts; over a time interval $\Delta t$ the dissipated <energy> is $P_{\rm loss}\Delta t$.