You're asking how long does it take for the sun to set (from lower limb to upper limb) as a function of time of year and latitude.
The simple way to answer this would to be run software that gives the altitude of the sun as a function of location and date/time. I have such software, but why use software when you can solve the problem analytically. It's much more fun that way.
If you don't like math, you can skip to the results below.
h = altitude above horizon
δ = declination of an object
φ = latitude of observer
H = hour angle of an object
First of all, there's one key equation:
The altitude of an object above the horizon is given by
sin h = sin δ sin φ + cos δ cos φ cos H
By setting h=0 in eqn 1, we can calculate the hour angle at which an object rises or sets:
cos H_set = - tan φ * tan δ
I'll make two assumptions here to simplify the calculation. First of all, the official definition of sunset is when the center of the sun is 50 arcminutes below the horizon. (That's 16 for the solar radius and 34 for refraction.) To make life easier, I'll assume that sunset occurs at an altitude of 0 degrees, not -0.833 degrees. This will have only a small effect on the result.
We can differentiate eqn 1 to get the rate of change of altitude over time:
dh/dt = (1 / cos h) (- K cos δ cos φ sin H)
The term K above is defined as dH/dt, the rate of change of hour angle as a function of time. Hour angle changes by 360 degrees in the course of one sidereal day (which is about 4 minutes shorter than a solar day). If we express K in terms of arcminutes per second, we have
K = 0.25068 arcminutes/sec
In the above equation, we've made our second simplifying assumption -- namely, we assume that the sun's declination is constant with time. In reality, it changes with time, but it doesn't change much over the time it takes for the sun to set.
We now manipulate eqn 3 as follows:
1) Since we are calculating this at sunset, we can assume that h=0.
2) The equation contains the term "sin H". Since we have a formula (eqn 2) for cos H at sunset, we can calculate sin H and insert it here.
3) Our goal is to calculate the duration of sunset (from lower limb to upper limb). The above formula gives the rate of change of altitude in arcminutes per second. To get the duration of sunset, we divide the angular diameter of the sun (32 arc minutes) by dh/dt. That will give us the duration of sunset in seconds.
Let T = duration of sunset
T = 32 / (K cos δ cos φ sqrt(1 - (tan δ tan φ)^2)) seconds
So that's our final equation -- the duration of sunset (in seconds) as a function of the sun's declination and the observer's latitude. The sun's declination is 0 at the equinoxes (March and September), and +23.4 or -23.4 degrees at the solstices (June and December). Note that the above equation gives the same result on reversal of the sign of either φ or δ.
At this point, one can set up a spreadsheet to calculate the above quantity for various parameters. I'll do the calculation for five different latitudes:
*** duration of sunset
latitude 0 (equator)
at equinox: 128 seconds
at solstice: 139 seconds
latitude 23.4 degrees (Tropic of Cancer)
at equinox: 139 seconds
at solstice: 154 seconds
latitude 33 degrees (San Diego)
at equinox: 152 seconds
at solstice: 173 seconds
latitude 40 degrees (Philadelphia)
at equinox: 167 seconds
at solstice: 195 seconds
latitude 52 degrees (London)
at equinox: 207 seconds
at solstice: 271 seconds
There are two obvious patterns here:
1) At each location, sunset is slower at the solstices than at the equinoxes.
2) Sunset gets progressively slower as you move away from the equator.
There's one other assumption I've made here -- namely, that the angular diameter of the sun is constant. In reality, the sun is closest in January and farthest in July; this effect alters the duration of sunset by up to plus or minus 1.7%.
A somewhat similar question is to ask about the duration of twilight (say, astronomical twilight) instead of the duration of sunset. Things are somewhat different here, because dh/dt is not constant as the sun sinks below the horizon, and there is no longer a symmetry between the solstices. This problem is left as an exercise for the reader.
I'll add some physical interpretation to all this. At the equinoxes, the sun moves in a great circle in the course of a day. At the equator at sunset, it moves vertically (perpendicular to the horizon), so the duration of sunset is short. At higher latitudes, it moves at an angle, so it takes a longer time to move 32 arcminutes vertically.
At the solstices, the sun's path during the day is not a great circle, but a small circle. The smaller the circle, the more slowly the sun moves through the sky. (For example, Polaris moves in a very small circle and moves extremely slowly.) Therefore, the duration of sunset is longer at the solstices (sun at declination +- 23.4 degrees) that at the equinoxes (sun at declination 0, on celestial equator).