Well, there is a simple model for heat pumps, based on conservation of energy.
Qhot = Qcold + W
where Qcold is heat moved out of the cold reservoir. W is work done on the system. Qhot is heat dumped into the hot reservoir, and due to conservation of energy, it turns out that, Qhot = Qcold + W.
For refrigerators, the goal is large Qcold in return for small W, so the figure of merit is,
COP = Qcold/W
where COP is well known acronym for "coefficient of performance". It is a measure of how good the fridge is at cooling.
For a heat pump as heater, the goal is large Qhot in return for small W, so the figure of merit is actually COP+1, since:
Qhot/W = (Qcold + W)/W = COP + 1
The people who make Peltier modules, publish their numbers for COP, and the exact number is variable, depending on the temperature difference across the module.
Search for "typical coefficient of performance for peltier modules"
returns a bunch of stuff. This page,
is kind of interesting, because it has a calculator to play with, and it does a good job of giving me an approximate number for COP for a (typical?) Peltier device. E.g. if I enter: 25 C for cold side, and 50 C for hot side, it says Peltier can give me COP=0.596 (and COP+1 = 1.596)
Also I found a blurb on this Wikipedia page,
"...the coefficient of performance of current commercial thermoelectric refrigerators ranges from 0.3 to 0.6, one-sixth the value of traditional vapor-compression refrigerators"
So if we go with Qhot/W =1.6, that is 1.6 times better than an ordinary (Qhot/W =1), cheap, electric heater.
Is that good enough for your thing? Does it justify the expense of using a Peltier module? I dunno. That's up to you, I guess.
BTW, I am just using that link to coolchips.gi for their calculator. I have no idea if the technology they're promising is real or not, or if they're just some sort of scam artists. I suspect the latter, but I don't know for sure.
Select as Best AnswerUndo Best Answer
1. All caps is seen as shouting - please don't the bold isn't making that better.
2. Raise it to what from 50 deg C?
3. What the 60 watts is desn't matter in theory but in practice you will have a lot of losses.
2.26 x 10 to the 6 joules per Kg. - this is to raise the water to boiling point.
you should be able to work out the rest. Sounds like homework.
Not only does it sound like homework again, the profile pic looks familiar too.