LM340 temp coefficient question

I have a question about temperature coefficients.
I'm using an LM340 (5 volt regulator) and don't understand how much heat it
will disapate and how much it will change over temp.
The datasheet specs 0.8mV/c but doesn't show a junction to case spec.
My questions:
Is 0.8mV/c positive or negative from what... 25 degrees C?
So if change my temperature 10 degrees C, my output voltage will be 8mv
higher (or lower) and from what external temp?
Also, how do I calcualte an accurate junction to case number?
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You perhaps missed the figures on the data sheet:
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Hint: The junction-to-case thermal resistance (?JC) for TO-3 and TO-220 package is 4°C/W.
The average temperature coefficient of output voltage is just that, an average slope. It is only at 0.8mV/c at one temperature (from the graph mentioned below, at about 60C).
Output voltage falls with increasingly positive temperature and with increasingly negative temperature, either side of its design temperature of 25C. The graph of "Output Voltage (Normalized to 1V at TJ = 25°C)" shows the effect.
See the graph mentioned above. It depends on the value of the initial temperature as well as the change in temperature.
No need, read the data sheet (more carefully, perhaps).
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