That looks right.

Also, given the equation q = m x C x ΔT

where q is heat (in Joules), m is mass in grams, C is the specific heat of the heatsink material in J/g*C or Joules/gram degree Celsius, and ΔT is the change in temperature, we can figure out how heat will affect our system (the laser).

For example (using your value for heat above):

Given a Joule is a Watt second (J = Ws)

Given the C of Aluminum is about 0.9

and assuming a heatsink mass of 10g

and assuming the heat at the diode is 4.88W

and assuming operation for 10 seconds

We get:

4.88W x 10s = 48.8J = 10g x 0.9J/g*C x ΔT

Solving for ΔT we get:

ΔT = 48.8/9 *C

The units (grams and Joules) cancel so our units are left in *C

So our change in temperature, then, after ten seconds is:

ΔT = 5.4*C

Since this value is positive, we can say that a laser diode producing 4.88W of heat into an Aluminum heatsink with mass of 10g will raise the temperature 5.4*C after ten seconds.

Note: ΔT = Temperature final - Temperature initial ; knowing this and using a little more algebra, if we also know the initial temperature (room temp, for example), we can determine how long it will take for the heatsink to reach a given temperature (Temp final) if we can't measure it directly. For solid objects, like a heatsink, it will not be easy to measure the temperature with a thermometer.

If you use room temperature as your Tinitial, be sure that your heatsink has reached thermal equilibrium with the surroundings.

Another note: This does not take into account radiation of heat from the system to the surroundings.

Another note: I don't have much physics training so if anything above is erroneous, please correct me.

For now I'll just be embarrassed in advance...