N-Channel Mosfet as Switch
Let's see how bad this looks...

According to the datasheet Vds max = 30V, what is much higher than 11.1V, so you are good in this regard.

Let's check the power dissipation @ Id=20A.
According to Fig. 12 of the datasheet, the Rdson degrades significantly when Vgs=5V instead of 10V. For Tj=125˚C (worst case), Rdson=16mΩ. So the power dissipated will be 0.016*(20A)^2=6.4W, what is pretty high.

According to the datasheet, the thermal resistance from junction to ambient with no heatsink is 62 ˚C/W (max). With Pd = 6.4W, assuming Tamb = 25˚C, we get Tj = 25 + 62 * 6.4 = 421.8˚C
!!! It's clear you need a heatsink!

Let's see how big of a heatsink you need. Let's say Tj=125˚C (what is already pretty high) and Tamb=25˚C, so the delta T will be 100˚C.
For Pd = 6.4W, the total thermal resistance will have to be less than 100˚C / 6.4W = 15.6 ˚C/W.
The thermal resistance from junction to the heatsink is 2.3 + 0.5 = 2.8 ˚C/W, according to the datasheet.
This means that the heatsink thermal resistance will have to be lower than 15.6 - 2.8 = 12.8 ˚C/W. You may be able to achieve that with a big heatsink or with a not-so-big heatsink with forced ventilation.

If you add a level shifter capable of driving the gate of the FET with 10V instead of 5V, the power dissipation will reduce significantly. According to the datasheet Rds max will become 8.7mΩ, what drops Pd to 3.48W, what will require a total thermal resistance of 28.7˚C/W, and 25.9˚C/W for the heatsink only. This translates into a much smaller heatsink.