Suppose we had two metal plates separated by an Angstrom. Now, apply a voltage $V$ between the two metal plates. There will be a tunneling current $I$ between the metal plates.
Since $P=IV$, does that imply power is being dissipated by the system, even if the tunneling is elastic?
Some additional thoughts: Perhaps this is related to why there is still resistance in ballistic transport.
Tunneling between two conducting regions separated by an insulator can be pictured in the following way (image source):
The voltage applied between the two sides of the structure amounts to shifting the chemical potentials of the two metallic regions in respect to each other (there is no voltage drop within the metallic regions themselves due to screening - at least, if the voltage is not changing too fast with time.)
Electrons from the region with a lower chemical potential cannot tunnel into the region with higher chemical potential due to the Pauli exclusion principle - since the states of the same energy are occupied. Indeed, the rate of tunneling from the state of energy $\epsilon$ in the right conductor to a state of energy $\epsilon'$ in the right conductor is given by a Fermi golden rule expression: $$ w_{R\rightarrow L}(\epsilon, \epsilon')=\frac{2\pi}{\hbar}|U_{RL}(\epsilon, \epsilon')|^2\delta(\epsilon - \epsilon')\rho_R(\epsilon)\rho_L(\epsilon')f_R(\epsilon)[1-f_L(\epsilon')], $$ where $f_{L,R}(\epsilon)$ are Fermi distribution functions. At zero temperature $f_{L,R}(\epsilon)=\theta(\mu_{L,R}-\epsilon)$, so that $$ f_R(\epsilon)[1-f_L(\epsilon)]= \theta(\mu_R-\epsilon)\theta(\epsilon-\mu_L) $$ The product of the Heaviside step functions is always zero, if $\mu_R < \mu_L$, since it differs from zero only for $\mu_R>\epsilon>\mu_L$.
The elasticity of tunneling is encoded in the delta-function of the Fermi Golden rule.
The current can be calculated as $$ I = 2e\int d\epsilon\int d\epsilon' \left[w_{R\rightarrow L}(\epsilon, \epsilon') - w_{L\rightarrow R}(\epsilon, \epsilon')\right], $$ where factor 2 accounts for spin.
Assuming that the matrix element $U$ and the densities of states $\rho_{L,R}$ are changing slowly with energy (so-called broad-band limit), we can treat them as constants and obtain: $$ I=2e\frac{2\pi}{\hbar}|U|^2\rho_L\rho_R^2(\mu_L-\mu_R)=\frac{4\pi e^2}{\hbar}|U|^2\rho_L\rho_RV=G_0|T|^2V, $$ where $G=G_0|T|^2$ is the the conductance of the junction. Note that on microscale the conductance/resistance do not necessarily scale with volume/length, and therefore cannot be expressed in terms of conductivity and resistivity.
The quantity $G_0=\frac{2e^2}{h}$ is called conductance quantum, whereas $T=2\pi U\sqrt{\rho_L\rho_R}$ is actually the first order approximation to the transmission matrix from the scattering theory. A more complete calculation shows that (without interactions) we could actually use here the full transmission matrix, as shown by Landauer and Büttiker. Thus, the conductance quantum gives us the conductance (and hence the resistance) of a junction when it is perfectly transmitting!
This seems paradoxical that we cannot have greater conductance (or lesser resistance) - indeed, in classical wire we could always increase the conductance by increasing the wire diameter, as $R=\rho l/A\leftrightarrow G=\sigma A/l$. The catch is that in considering current in one dimension we have completely ignore the transversal modes - given $n$ such modes we would get much higher conductance $nG_0$. Howe many modes are exactly included depends on the transversal potential and the magnitude of the bias, which is why quantum conductance does not linearly scale with size.
Where does the energy go?
Now, if the tunneling is elastic, but the conductance is finite, where is the energy dissipated?
The explanation is that this model assumes that the metallic regions are held at constant potential and are in thermodynamic equilibrium. The only way to maintain this equilibrium when a high energy electron arrives from (or when an electron leaves to) the other metallic region is for the electron/hole to relax quickly to the Fermi surface (i.e., the chemical potential). The relaxation process (e.g., via a phonon emission) is not included in our model - it is not a part of the tunneling, but this is where the energy is released (possibly in the form of Joule's heat.)