MOSFET heat calculation question
Your analysis oversimplifies a few things, but works out more or less correctly in the end. Just for general edification, I'll point out a couple things.

First, Rds isn't a fixed number, it varies with Vgs, temperature, and current. Figure 1 of the datasheet shows Vds vs. Ids for various values of Vgs. The reciprocal of the slope of this curve at a given current value is Rds.

For Vgs of 5V, the graph doesn't show anything for Ids = 2A, but we can extrapolate. Picking two points on the Vgs = 5V curve, we get a slope of 30 amps per volt, or (phrasing another way) an effective Rds of 33 mOhm.

Note that this assumes a junction temperature of 25C. From figure 2, we can see that at low currents, a hotter junction actually results in an even lower voltage for a given current, translating to a lower Rds. If you were running in a sub-zero ambient (say, equipment that lived outdoors in February in Moose Jaw), your Rds would be higher. But let's assume you're indoors in standard room temperatures.

So two amps RMS through 33 mOhm dissipates 132 mW of power, P = I^2 * R. .132W x 62.5 K/W gives a junction rise above ambient of 8.25K, or 14.85F. (1K = 1.8F) You might notice it get a little warmer to the touch, but that's it.

Max operating junction temperature is listed as 175C, so as long as your ambient is below 166.75C, you'll be fine thermally. If your ambient is higher than that, you're doing something either wrong or awesome.