A soapbox is hardly a standard unit of measurement, but 12 W doesn't require anything too huge, even without forced air, provide you can at least let natural convection be, well, natural. Here's how you calculate the heatsink you need.
I've picked the datasheet for IRF510 as an example. It's a very common TO-220 MOSFET and should work for your application.
The first thing you will see is that the datasheet lists power dissipation as 43 W. This of course requires an ample external heatsink, but it should cover your application with a healthy margin.
The absolute maximum junction temperature \$T_J\$ is listed as \$175 ^\circ C\$, and let us assume ambient temperature is \$35 ^\circ C\$. That means the temperature can't rise more than \$175 ^\circ C - 35 ^\circ C = 140 ^\circ C\$. And to be safe, let's add a safety margin and design for no more than \$ 100 ^\circ C \$ rise.
The datasheet lists the maximum junction-to-case thermal resistance as \$ R_{\theta UC} = 3.5 ^\circ C/W \$. That is, for every watt, the junction temperature will rise \$ 3.5 ^\circ C\$ assuming the heatsink can magically remove all heat. At 15 W, that's a rise of \$ 3.5 ^\circ C/W \cdot 15 W = 52.5 ^\circ C \$.
We are hoping for no more than a \$ 100 ^\circ C \$ rise, so we will have to find a heatsink that won't raise the temperature more than another \$ 100^\circ C - 52.5 ^\circ C = 47.5 ^\circ C \$. That means our thermal resistance budget for the heatsink is \$ 47.5 ^\circ C / 15W = 3.17 ^\circ C / W \$.
This is pushing the edge of what can be done with natural convection, but it's doable. Thumbing through my Mouser catalog I can find an Ohmite heatsink FA-T220-64E with a thermal resistance of \$ 3 ^\circ C / W \$ with natural convection. It's the biggest one they sell for TO-220. It's about 1 x 1.6 x 2.5 inches and Mouser will sell you just one for $2.17 plus shipping.
Strictly speaking, I haven't taken into account the thermal resistance of the transistor case to the heatsink. The IRF510 datasheet gives a typical value of \$ 0.5 ^\circ C / W \$ for a greased surface, which at 15 W will mean another \$ 7.5 ^\circ C \$ rise in junction temperature. But remember we included a margin of \$ 40 ^\circ C \$ and assumed a rather high ambient temperature of \$ 35 ^\circ C \$. We should be safe.
Even so, something may obstruct your heatsink, so you may do well do build in some sort of thermal protection. You can implement this yourself, but there are also MOSFETs out there with thermal protection built in. If you do this, you don't need such a margin, and you may very well be able to dissipate more than 15 W if you don't mind the possibility that the thermal protection kicks in.
And, it bears mentioning that even though the transistor shouldn't fail, it will get mighty hot. A plastic box with poor ventilation is probably no good. You will have to keep fingers away for sure. If you want to keep things cooler I'm afraid you have no choice but forced air or spreading the heat over more area: big power resistors, multiple transistors, etc.
Best Answer
According to the datasheet it can dissipate up to 0.5W at 50˚C ambient temperature, which would make the maximum temperature about 150˚C. This leaves plenty of room for derating in long-term use, so you shouldn't have problems. Additionally, you could keep the leads short and use the PCB for a bit of power dissipation.