Energy cost of adiabatic quantum computation
Basically, the computation is reversible, so you are not actually using energy. However, the rate at which a computation progresses is generally proportional to the energy gap within the system. In the case of adiabatic quantum computation, the rate at which you can transition between ground states of the initial and final system is determined by the energy gap between the ground state and the low lying excited levels of the system. Thus the energy gap in the Hamiltonian (the total energy operator) is what determines the rate of computation, and so scaling up the energy levels involved allows the computation to be performed more quickly, even though the process does not actually consume the energy.

You might be interested in looking at the Margolus-Levitin theorem, which gives limits for the rate of computation by imposing a lower bound on the time taken for a system to transition to an orthogonal state in terms of the energy gap present in the system. This applies more generally than simply to adiabatic quantum computing.

In general, you tend to get a linear trade-off between energy scale and time. One way to see where this comes from is simply to think about the case of a constant Hamiltonian. In that case, the time evolution operator is $e^{-iHt}$. Clearly multiplying $H$ by some constant $k$ and dividing $t$ by $k$ leads to the same operator, and hence the same evolution of the system. You can do this more generally, with a potentially time dependent Hamiltonian by looking at the Schroedinger equation: $i \hbar \frac{\partial}{\partial t} \mid \Psi \rangle = H \mid \Psi \rangle$. Replacing $H$ with $kH$ and $t$ with $\frac{t}{k}$ we obtain the same equation.