Well no not to any desired temperature, but you can estimate within engineering accuracy what the temperature rise will be by dimensional analysis. It is not an exact solution but will be accurate within an order of magnitude.
For this problem it is quite trivial but for more exotic equations (perhaps with heat transfer to the surroundings, or 2D etc.) dimensional analysis is a powerful tool which can provide insight into the dynamics of an equations which may otherwise be unsolvable (e.g. due to non-linearity, etc).
First, lets state the system we are dealing with:
$$\partial_{t}T=a\partial_{x}^{2}T\quad T\left(x,0\right)=T_{0}$$
with thermal diffusion coefficient $a=k/\rho c_p$ and boundary conditions:
$$-k\partial_{x}T\left(0,t\right)=Q\quad\partial_{x}T\left(L,t\right)=0$$
Now for the dimensional analysis we will determine characteristic time ($\Delta t$), length ($\Delta x$) and temperature ($\Delta T$) scales which will result in the above equations to only contain terms of $O(1)$ and smaller when they are non-dimensionalized.
We define the relevant dimensionless variables:
$$\theta=\frac{T-T_{0}}{\Delta T}\quad\eta=\frac{x}{\Delta x}\quad\tau=\frac{t}{\Delta t}
$$
Substituting these dimensionless variables in the equations yields:
$$\partial_{\tau}\theta=\frac{a\Delta t}{\Delta x^{2}}\partial_{\eta}^{2}\theta\quad\theta\left(\eta,0\right)=0$$
$$-\partial_{\eta}\theta\left(0,\tau\right)=\frac{Q\Delta x}{k\Delta T}\quad\partial_{\eta}\theta\left(\frac{L}{\Delta x},\tau\right)=0$$
To obtain equations where all terms are $O(1)$ or less we define:
$$\Delta x=L\quad\Delta t=\frac{L^{2}}{a}\quad\Delta T=\frac{QL}{k}$$
which result in the equations becoming:
$$\partial_{\tau}\theta=\partial_{\eta}^{2}\theta\quad\theta\left(\eta,0\right)=0$$
$$-\partial_{\eta}\theta\left(0,\tau\right)=1\quad\partial_{\eta}\theta\left(1,\tau\right)=0$$
which are indeed $O(1)$ or smaller.
Having determined the characteristic scales of the system, it follows that it take approximately time $\Delta t = L^2/a$ to increase the temperature by $\Delta T = \frac{QL}{k}$. Hopefully, you find this useful.
