Heat dissipation on high power resistor
There is no simple formula. The result will depend on how the copper plate is positioned relative to gravity field, horizontally or vertically, is it really in a free air, or if there are obstacles to airflow.  And the shape of copper plate (aspect ratio) will matter.

To get some ideas of reasonable values of thermal impedance of various sink geometries, take a look at this article, "Calculating heat sinks"

You are actually asking to dissipate 50 W with delta_T of 10 deg.C. This is equivalent to thermal impedance of 0.2C/W. This kind of impedance can be achieved only with multi-fin sink design under forced air, like a CPU heat sink. The value of 0.2 C/W can't be achieved with a simple copper plate in free air, maybe a massive heat spreader based on efficient heat pipes can help, but I doubt it.

Today the design of heat management configurations is usually done using expensive software such as FloTERM,  which uses solid thermal models for heat conduction, and associated airflow models for convection around the solid surfaces. It is pretty complicated, and thus expensive.