Just as a quick thing... max temp for a crock pot is a very sticky animal...
Beware, massive geekery ahead!
Power is a measurement of applied energy, a watt is one joule of energy applied per second. We have the wattage of our appliance (and a way to vary that wattage between a maximum and a minimum).
But how do we convert that to temperature change?
The British Thermal Unit (or BTU) is defined as as the energy required to raise one pound of water, one degree farenheit (those of you who have had recent experience with a water heater or air-conditioner/furnace will recognize this term). Well, lets convert this to a metric unit so we won't have to muck around with it later.
We know that 100*C x 9/5 + 32 = 212*F which means that a 1*F change is 5/9ths of a 1*C change. So the Evolved unit is "energy to raise one pound of water 5/9*C".
And we know that 1 pound equals .4536 kilograms. So the Evolved unit is "energy to raise .4536 kilograms of water 5/9*C".
So our Metric Thermal Unit (which is directly equivilent to a BTU) is the energy required to raise .4536 kilos of water 5/9ths of a degree Centigrade.
Well, now we need to know the volume of what we're cooking. Lets say we have a 4 quart pot, and a standard load of 3/4 full. Which works out to 3 quarts. One quart is about .9464 liters, so 3 quarts is roughly 2.84 liters. Lets convert that to mass of water (as water is probably the most abundant thing that's going into the pot) as 2.84 kilos.
Lets figure on a final temperature of about 175*f or about 80*C and a starting temperature of about 65*F or 18*C (figures agressively rounded). So we need to change our mass 62 degrees C.
So we need to change 2.84 kilos of water 62 degrees C.
2.84 kilos / .4536 kilos = '6.26 blocks' of water to be heated. Each of which must be heated by 62*C / (5/9*C) = 122.4 "units" of temperature. Therefore we need 766.224 MTUs (applied evenly throughout the mass) to
instantly heat our water to desired temperature. Or that amount of MTUs applied over time (neglecting any cooling effect).
So what's a good time frame to hit 170ish F? Well, we don't want to hit 170 at the end of the cooking cycle, and we don't want to hit it too early. So lets say we want to hit 170ish in 2 hours. Which means we have to apply 766.224 MTU over two hours... or about 383 MTU/Hour.
Thankfully we have a handy little constant that will convert from our MTU (which is effectively a BTU) to watts... One BTU/hour is equal to .293 Watts. Which means 383 MTU/hour x .293 = 112.219 watts.
So we hit 170ish in two hours at 112 watts (neglecting cooling). Thing about that is, at that rate we hit a temperature change of 62*C every two hours... which means that by the end of the cooking time we'll have exceeded 500*F... and we know that doesn't happen (in a crock pot at least).
The thing is, that most crock pots don't operate by thermostat, they operate by continually putting heat into the pot, and the principle that heat moves from hot objects to cooler objects. So the heat generated in your crock pot moves through the walls and into the air. When heat in equals heat out, your crock pot is at max temp.
Which means that the absolute Max Temp that your crock pot will reach is not determined by the wattage, but by exactly how fast heat escapes from your crock pot. This changes depending on exactly how hot the crockpot is, what it's made of, air temperature, airflow around the crock pot, and a few other factors.
So it's a balancing act. We want enough heat going into our pot to raise the temperature to our desired levels, but we don't want to put so much heat into the pot that the level of heat gets too high (which is what it's doing as a regular thing). So we need to reduce the wattage used, but not by too much.
When you reduce the wattage used by the appliance, you don't lower the max temperature so much as change the balance point of energy added vs energy lost, because the rate that heat moves out of an object changes dependant on the difference in temperature between the object and it's environment (or trying to use a crockpot outside in the snow will not give you the same results as one used in a sauna).
I expect if you plotted the max temp vs wattage for any individual pot, you'd wind up with something that looks more like an exponential function, rather than a linear one, so we can't use a direct relation between wattage and max temperature.
"Enough about heat transfer, how does this apply to my crock pot?" you ask? Well, if you want a functional answer, rather than a mathematic one, you use what Keltin already thought up. Get a dimmer (Keep in mind that this dimmer is going to be getting pretty hot it's own self, it's basically just a big variable resistor, and when you run voltage through resistance, you get power expressed as heat) and set it so you get the right wattage in your pot. But where to start?
Keltin started by using a voltage example. If you wanted the dimmer to use 1/4 the voltage (and leave 90v left for the crock pot) you would set it to approx 1/3d the resistance of the crock pot. So if your crock pot was 60 ohms you would set your dimmer to about 20 ohms. You'd have a total resistance of 80 ohms. You would be putting through a current of 1.5a. 120v/80 Ohms = 1.5 ampres.
Lets take that info and find the wattage developed by the pot. Twinkle twinkle little star, power equals I-squared R. or P = I^2 x R.
(1.5a)^2 x 60 ohms = 135 watts.
Well lets see what we need to set our dimmer at to develop 120 watts (our figure of 112w above, plus 8 watts to negate cooling) in our pot.
120 watts, divided by 60 ohms equals the square root of 2 amps, or 1.414.
To get 1.414 amps flowing from 120 volts. we need a total resistance of 85 ohms. That means we have to set our dimmer at 25 ohms to get the total resistance (Don't forget, your dimmer will be developing (1.414^2) x 25 = 50 watts of power itself). The fact is though, 120 watts is probably way low to actually be a viable slow cooker wattage, simply because of all the real world losses that come into play. I'd start somewhere closer to 130-135.
Now if you need more heat, (and if you use 120w as a starting figure, you probably will) you lower the resistance of the dimmer (which raises current, and wattage deveoped by the pot.)
So get your pot, measure the resistance of the coil on low. Call that figure 'C'.
Now divide your desired wattage (130 is probably a good place to start as it's probably lower than what you have now, but isn't so low that you're gonna have to work back up). Call your desired wattage 'D'.
Divide D by C and take the square root of the result (for example using D= 130 and C=60 gives us a result of 2.17 or an answer of about 1.47). Call this number 'I'
Divide 120 by I (voltage of the circuit by current) to get total resistance, then subtract C. (for I=1.47, 120/1.47= 81.63, that minus 60 is 21.6 (or 22 is close enough...). Call this number 'S'.
Grab your multimeter and set your dimmer for your value of S. Fill your crock pot to a normal level (basically with about how much mass you usually cook, Water is your best bet, as you can't be sure exactly how well this will cook), plug your contraption into the wall, set to low and wait for about 3 hours. Measure temp. Wait another hour or two, measure temp. If they were both close to the same and where you want them to stay during cooking, mark your dimmer and you're done with low. If not, choose a new wattage, run the calculations and try again.
Then repeat for high.
And now I'm done.
