Why would MOSFET transistor heat up?
Mirroring what Brian Drummond said above, it may not be switching on 'very hard'.  What this means is that the MOSFET is not entering the 'active' region of operation.  In the 'active' region, you can generally expect the MOSFET to have an \$Rds\$ that matches the specified \$Rds_{on}\$ (in the case of the BUZ11, 0.040 ohms).

What is instead happening if the MOSFET is not in the active region is that it is in the 'linear' region of operation.  The conditions to be in either of these two regions is:


linear region: \$V_{GS} &gt; V_{th}\$ and \$V_{DS} &lt; ( V_{GS} – V_{th} )\$
active mode: \$V_{GS} &gt; V_{th}\$ and \$V_{DS} ≥ ( V_{GS} – V_{th} )\$


(The above from This SO answer)

If you are operating in the linear region instead of the active region, your resistance between drain and source is going to be greater than the specified \$Rds_{on}\$. for that part; even if your \$V_{gs}\$ is above the specified \$V_{th}\$.

They key point in this is knowing that the gate-source voltage is not enough to guarantee that the resistance of the MOSFET is going to meet the specified on-resistance on the datasheet.

If you have higher resistance than you expected, that might affect your power calculations; e.g. you might be dissipating more heat in the MOSFET than you otherwise expected.