Let's say I have 2 identical metal bars, same properties and everything, both are at 100 degrees Celsius. Is the time heating up one metal bar to 120 the same as cooling one down to 80 degree Celcius?
I think we can assume that it is, formula for heating/cooling of metals is $$Q=m*c*\Delta t$$ where $t$ is temperature, as $\Delta t$'s are equal, same with required heat and time if we use same power in both cases.
Expanding on the comments...
One of the ways to figure that out without solving equations is to take advantage of free and easy to use electrical spice programs and model heat transfer with electrical circuits. This should work because, with a proper modeling, since underlying equations for charge and heat transfer are similar.
Let's say, we want to set up symmetrical cooling and heating conditions (as proposed in the comments), using identical bars as cooling and heating bodies.
The metal bar at 100 degrees could be modeled as a capacitor, say, 1uF charged to, say, 100V.
The cooling bar with the initial temperature of 60 degrees could be modeled by a 1uF cap charged to 60V. Similarly, the heating bar at 140 degrees could be modeled by a 1uF capacitor charged to 140V.
The thermal resistance of the thermal contact between the bars could be modeled as an electrical resistance between the capacitors, presumably, the same for both cooling and heating scenarios, say, 1Mohm.
The circuits and the simulation results are shown below. Here we can see that, due to symmetrical conditions and linearity, the cooling and heating will take the same time.
NOTE: Since we don't know the actual physical characteristics of the bars, we are picking random electrical equivalents and, correspondingly, the results of the simulation should not be interpreted as absolute.
If, for cooling and heating, we chose massive bodies which would maintain constant temperatures throughout the process, we could model them as perfect voltage sources of 80V and 120V, where 80V source would act as a heat sink and 120V source would act as a heat source.
Here, again, due to symmetrical conditions and linearity, the heat transfer will take the same time.