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Heat sink ratings
Heat sinks are rated by their thermal resistance (Rth) in °C/W. For example 2°C/W means the heat sink (and therefore the component attached to it) will be 2°C hotter than the surrounding air for every 1W of heat it is dissipating. Note that a lower thermal resistance means a better heat sink.
Working out the required heat sink rating:
First find the thermal power to be dissipated:
P = IC × VCE
(if in doubt use the largest likely value for IC and assume VCE is half the supply voltage).
Example: a transistor is passing 1A and connected to a 12V supply so the power is about 1 ×  × 12 = 6W.
Find the maximum operating temperature (Tmax) for the transistor if possible, otherwise assume Tmax = 100°C.
NPN transistor with load
Estimate the maximum ambient (surrounding air) temperature (Tair). If the heat sink is going to be outside the case Tair = 25°C is reasonable, but inside it will be higher (perhaps 40°C) allowing for everything to warm up in operation.
Work out the maximum thermal resistance (Rth) for the heat sink using:
Rth = (Tmax - Tair) / P
With the example values given above: Rth = (100-25)/6 = 12.5°C/W.
Choose a heat sink with a thermal resistance which is less than the value calculated above (remember lower value means better heat sinking) for example 5°C/W would be a sensible choice to allow a safety margin. A 5°C/W heat sink dissipating 6W will have a temperature difference of 5 × 6 = 30°C so the transistor temperature will rise to 25 + 30 = 55°C (safely less than the 100°C maximum).
All the above assumes the transistor is at the same temperature as the heat sink. This is a reasonable assumption if they are firmly bolted or clipped together. However, you may have to put a mica sheet or similar between them to provide electrical insulation, then the transistor will be hotter than the heat sink and the calculation becomes more difficult. For typical mica sheets you should subtract 2°C/W from the thermal resistance (Rth) value calculated in step 4 above.
Or use trial and error!
If the steps above seem too complex you can try attaching a moderately large heat sink and hope for the best. Cautiously monitor the transistor temperature with your finger, if it becomes painfully hot switch off immediately and use a larger heat sink.
Why thermal resistance?
The term 'thermal resistance' is used because it is analagous to electrical resistance:
The temperature difference across the heat sink (between the transistor and air) is like voltage (potential difference) across a resistor.
The thermal power (rate of heat) flowing through the heat sink from transistor to air is like current flowing through a resistor.
So R = V/I becomes Rth = (Tmax - Tair)/P
Just as you need a voltage difference to make current flow, you need a temperature difference to make heat flow.
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