My gut feeling is that using that kind of battery to create any more than a hand warmer is not very practical. To get an estimate if the idea of a battery powered heater is doable you have to look at the energy needed:
When heating things the energy is used to do two things:
raising the temperature: to heat 1kg of water from 30°F to 60°F you need the following energy: (4180 J/K) * 16.7K = 70KJ (heat capacity times temperature difference in Kelvin). For reference: An alkaline AA battery has about 9360J (source). You can substitute your own data for e.g. your pipe (water filled?) combination to get the result for your application.
replacing the heat that is lost due to bad insulation: If e.g. your pipe is warmer than the ambient temperature it will get colder. Here you can also calculate the energy that is lost this way: P = (Temperature difference) / (thermal resistance) with P being the heat lost measured in W (= J/s)
I would start with plugging in the numbers for part 1 and then add an estimate for part 2 ("if I use insulation X, how much energy would I need to have the temperature be constant for time X?"). Then compare the energy needed with the amount stored in the type of battery you want to use.
I think you can assume pretty much 100% efficiency, but better have the batteries be warm or they will not deliver their full power.
It probably is feasible.
But: you need to realize that even for electricity, conservation of energy applies! So if you want to produce 10,000 Ws (wattseconds) = 10 kJ of heat, you need to have that much energy in your battery.
So, in your place, I'd start by calculating what amount of heat you want to produce. That's really easy. Calculate the amount of material (ie. flesh) you want to heat up, multiply that by the specific heat of the material (use the specific heat of water, flesh is mostly that stuff), and then you know how many Joule you need. One Joule is 1 W · 1 s.
Then divide that by the time (in seconds) you want to allow the heating to heat up that much material, and you've calculated the power (in Watt) necessary to heat up that amount of material that fast.
Now, I don't believe you know nothing. You've probably heard of 1 W = 1 A · 1 V. So if you have a 12 V battery, in order to produce 1 W, needs to make \$\frac1{12}\text{ A}\$ current flow. If you need 120 W, you'll need 10 A. It's that simple.
Now that you know how much current you need to spend, you can calculate two things:
- how long your warmer should last with a single charge. Most batteries are rated in "Ampere hours", Ah, so take one battery that you think would be of acceptable weight and size, and figure out how long it will be able to source that current.
- how much resistance your wire needs to have, because, Ohm's law, U = R · I -> R = U / I (the resistance is the ratio between voltage U=12V and the current I). Heater wire has a fixed resistance per length. You can thus calculate the length of wire necessary.
This doesn't address things like maximum power impedance matching etc, but it does in theory give you a good idea of what your system needs to look like.
Best Answer
No, you don't need an isolation transformer provided the low voltage power supply, relay and temperature sensor are isolated from the resistor and mains circuit. If any of those isolation barriers is not present or fails, even for an instant, bad things may happen (such as a shock, fire, destruction of your computer, etc.)
The power the resistor can dissipate is given in the datasheet.
If you are using it to heat something, say to 100°C, then you can dissipate about 80% of 50W or 40W. The actual power will depend on how insulated the resistor is as well as the ambient temperature (it's the air immediately around the resistor that matters).