Note that there's a short summary and a guide linked to at in the second to last paragraph! Otherwise, enjoy the massive dump 'o information!
So unfortunately the reason you can't find anything which directly relates power to range is because the answer is really, really complicated and depends on a lot of factors. If you assume enough aspects you can simplify the equations down to a ballpark estimate, but even that depends on a lot of factors.
So to start, the power measurement you're given is the amount of radiated power delivered to the connector which will attach to your antenna. This is then distributed by the antenna you attach, but the radiation pattern of the antenna you attach will make a big difference. This is because the power is not equally distributed in a spherical pattern, it will be directed in some fashion. You can see a bit of what the analysis looks like on this wikipedia entry.
Looking at the picture you provided, that's likely either a duck antenna (less likely) or a vertical dipole. Dipoles like that have a nearly-spherical radiation pattern with conical gaps at either end, and if they're vertically oriented have the benefit of being omni-directional (excluding up and down, which are normally not that important). The downside to this is that vertically polarized signals are more likely to be absorbed rather than reflected off the surface of the earth, thus limiting their range. But I digress.
This means that the dipole will lose power relative to the surface area they cover at a given distance, plus the loss to the atmosphere absorbing the signal. The easiest way to think about this is imagine that you had a light bulb in the middle of a dark field. If you held up a white sheet of paper 100 feet away, it would hardly be lit, even with a very bright bulb. If you stood right next to the bulb, it would be bright! In the same way, radio signals will be weaker the further they are from the source. This dissipation will happen in accordance with the inverse-square law. So a more directed signal will be longer range because the initial power would be spread over a smaller surface area, but you have to be in the way of its beam, which limits the application.
So, got all that? You take the power, calculate the radiation pattern and where you expect the receiver to be, and that will give you a mW per unit area at the receiver. Great, right? Now you have to follow that up with the receiver!
Pretty much the same whole bunch of ridiculous stuff comes up on the receiving end: you have an antenna which will have certain characteristics and those characteristics will guide the reversal of an input signal at a given power to a corresponding voltage entering the electronics. Then the electronics will have a bunch of requirements about the voltage level it will be able to distinguish relative to the noise floor (i.e. how many other things are radiating nearby in the same frequency), and that will tell you if you can or can't reliably receive.
Long story short? An exact range is very difficult to give, and for many people, the answer wouldn't apply. The ballpark ranges depend on your application, receiver sensitivity, and antennas. While researching this answer I found a guide which uses something called the Friis transmission equation which I've never heard of, but it looks like a great approximation. The alternative is to find a transmitter which you already have, and you know the power of, and see at what distance it stops working as you would like it to. Then apply the inverse-square law to see how much power was at the distance you were standing. Then take the new transmitter power, and see what distance you would need to stand at to get the same power to your receiver.
Hope that helps! Sorry for the massive dump of info.
Edit: And just to have it in there, frequency scales the distance as well. Higher frequency = shorter distance is the short version. As you can see in the Friis equation, it looks like it scales proportionally to the inverse of the frequency.
dBi is defined as the ratio of the radiated power of an antenna at its maximum response angle to that radiated by an isotropic (hence the i) antenna, measured in dB. Thus if an antenna radiates twice as much power at its maximum response angle than an isotropic antenna, when the input power to both antennas is the same, than that antenna is said to have a gain of 3 dBi. Isotropic antennas radiate equally at all angles. For most practical antennas, the radiated power varies with angle. This variation is defined as the radiation pattern of the antenna. However, the dBi gain is defined only at the maximum response angle of the antenna so it does not vary with angle. It is quite possible for two antennas to have the same dBi gain, but very different radiation patterns.
Best Answer
The standard link loss equation is: -
Link loss (dB) = 32.4 dB + 20log(MHz) + 20log(kilometres)
If your RF sensitivity is -121dBm (about right for 1 kbps) and you can transmit at +20 dBm then it could theoretically (in outer space with very little interference and without earth getting in the way) work with a link loss of 141 dB (20 dBm - (-121 dBm))
So, we now have 141 dB = 32.4 dB + 59.2 dB = 20log(km)
Therefore 20log(km) = 49.4 dB which means distance = 295 km = 183 miles.
All sounds great until you factor in real-world problems such as fade margin - this is a kind of rule of thumb that suggests that link loss (at any given distance) is usually degraded by at least 20 dB. I'm not going to justify this BTW.
This now means distance is 29.5 km = 18.3 miles.
By the way, 0 dBi antennas usually imply isotropic antennas (a theoretical device) but straight wires usually imply a quarter wave dipole and these have a small gain of about 1.7 dB.
If you were transmitting at 28 kbps, your receiver sensitivity will be decreased to: -
154 dBm - 10log(data rate) = 109 dBm - this reduces your range to about 7.5 km = 4.6 miles.
If you are in a built-up area with buildings and other forms of interference you might lose another 10 to 20 dB so be aware of this.
Here is a very approachable document that largely covers what I've said in my answer.