Temperature rise is something you have to consider, but usually the resistance and the resulting voltage drop at full current have been the limiting factors when I've gone through this. That said, 100°C is a large temperature rise. That's not enough to be a problem for a copper trace on a FR4 board by itself, but that's going to affect the apparent ambient temperature for nearby components.
If you have that much temperature rise, you're dissipating significant power in the trace, which means power loss in your system. Again, the first concern should be how much voltage drop you can tolerate. Once you get that to acceptable levels, the temperature rise is usually low enough.
Also consider that 2 oz copper and more is widely available. The extra cost of specifying 2 oz copper for outer layers may be less than making the board larger or dealing with the heat or voltage drop. 2 oz on outer layers doesn't usually add that much cost. If you stitch together a trace on both outer layers, you have 4x the copper cross section than for a single trace of 1 oz thickness. If it's only one or two traces in a otherwise low current design, you can leave the soldermask off the trace and have a copper wire soldered over the trace. There are actually bus bars meant for this. However, consider the manufacturing cost. 2 oz copper may start to look like the cheap option when you consider the total cost of alternatives.
Again, look at all the options and all the criteria for deciding on trace width. Don't just focus on temperature rise, or assume that thicker copper is more expensive once the whole system is considered.
I'm going to have a stab at some maths :)
The DC resistance of a conductor - any conductor - is calculated as:
\$R_{DC} = \frac{{\rho}l}{A}\$
Where \$\rho\$ is the resistivity of the conductor in \$\Omega/m\$, \$l\$ is the length in meters, and \$A\$ is the cross-sectional area in m².
The thickness of 1oz copper is \$0.000034798m\$. Say you have a 3mm (or 0.003m) wide trace. The cross-sectional area is (approximately, assuming a perfectly rectilinear cross-section) \$0.000034798 × 0.003 = 0.000000104m^2\$. Resistivity of copper is \$1.68×10^{−8}\$ at 20C, and your trace is 100mm long (0.1m).
\$R_{DC} = \frac{1.68×10^{−8} × 0.1}{0.000000104} = 0.016153846\Omega\$ at 20C.
Ok, now for the tricky bit. The temperature co-efficient (\$\alpha\$) for copper is 0.003862.
\$R(T) = R(T_o)(1+\alpha{\Delta}T)\$
So for a temperature of 30C we have a \${\Delta}T\$ of 10C, or 10K (30 - 20 = 10, K = C + 272.15).
So \$R(30) = R(20)(1+0.003862×10) = 0.016153846×1.03862 = 0.016777708\Omega\$
So now solve Ohm's Law for voltage. Say you have 100mA flowing through the trace. That's \$V=RI\$, so \$0.016777708×0.1 = 0.001677771\$ or \$1.678mV\$ dropped across the trace at 30C.
Who says you need online calculators?
(Now, it's been about 20 years since I did this kind of thing at college, so I may be completely wrong ;) )
Best Answer
If you have equations for (say) 1oz (34.79um) copper, simply pre-scale the equivalent current based on the resistivity and trace thickness.
If the trace is carrying 0.5A and the material has bulk resistivity 3 times that of copper and it's 60um thick, then it's the same as a copper trace carrying 0.5 * sqrt(3) * sqrt(34.79um/60um) = 0.66A.
Heat loss should not be affected significantly by trace thickness or resistivity.