I know this is an old thread, but I discovered it researching this subject and wanted to correct/add a couple of things.
The formula to find the required thermal resistance of the heat sink given by jluciani is basically correct but is missing a term for the ambient temperature (Ta). The equation should be:
Tj = (Rjc + Rcs + Rsa) * Pd +Ta
Where Tj is the maximum target temperature of the junction. I will use 125 degC as the maximum temperature of the junction to allow a safety margin in the event the ambient temperature goes over the standard 25 degC. This gives:
125 = (1.92 + 0.5 + Rsa) * 36 +25
Rsa = (125-25)/36 - 1.92 - 0.5 = 0.3577 degC/W
The next part to find the size of the aluminium plate required to achieve this low a thermal resistance is much more complicated, but this blog https://engineerdog.com/2014/09/09/free-resource-heat-sink-design-made-easy-with-one-equation/ gives a very simple rule of thumb aproximation given by:
Area = (50/Rsa)^2 cm2
Unfortunately this formula applies to passive heat sinks with fins and I believe the author made a typo and meant area = 50×(1/Rsa)^2. The fins make a big difference. After looking at the results of of this online calculator https://www.heatsinkcalculator.com/free-resources/flat-plate-heat-sink-calculator.html and the data sheets from a range of passive heat manufacturers I did a bit of curve fitting and came up with this more comprehensive ball park formula:
Area = (20*1/(1+flow)*1/(0.25+h)*1/Rsa)^2 cm2
Where flow is any flow from a cooling fan in cfm and h is the height of any fins.
For the situation in the OP there is no forced cooling so flow= 0 and there are no fins, so h = 0 and the formula simplifies to:
Area = (80/Rsa)^2
Given that we require a thermal resistance <= 0.3577 the size of plate required to cool the transistor in the OP is:
Area = (80/0.3577)^2
= (223.6 cm)^2
This is probably too large to be practical.
As Kevin Vermeer pointed out, this particular transistor in this service is not really suitable for passive cooling. However, a dramatic decrease in the heat sink size can be obtained by adding fins and a fairly modest cooling fan as shown by the chart at the bottom of this link
Staying with a flat plate and adding a fairly good PC cooling fan of 100cfm air flow, the plate size could be reduced to:
Extruded aluminium can be bought in long strips with fins and using such a finned plate with 3cm fins and no cooling fan would require a heatsink size of:
Area = (20*1/(0.25+3)*1/0.3577)^2
Finally, combining forced cooling of 100cfm and 3cm fins gives:
Area = (17.2/(1+100/8))^2
Pressure drops and proximity of other hot components in the cabinet can reduce the efficiency.
Dust ingress can insulate heat sinks and cause cooling fans to slow down and fail over time.
Heat sinks that are much much larger than the contact area of the component they are cooling loose efficiency due to the distance the heat had to travel to spread to the extremities of the heat sink
Follow the usual guidelines on ensuring good contact with component to be cooled using a thin layer of a suitable heat transfer compound between the contact surfaces.
Results from this formula for extremely small or large heat sinks should be treated with suspicion. For example in the last result the cooling fan radius is much larger than the heat sink and so most of the airflow would not be flowing in close proximity to the fins and so the result is suspect. Otherwise, it is a pretty good approximation.
Probable best to add 25 degrees to whatever you think the ambient air temperature is and deduct a 25 degrees margin of safety from the maximum target temperature of the component when carrying out the calculations, just to be on the safe side.
Dont use this formula to design the cooling for a nuclear power station.