How to Generate Heat with Joule Effect? [closed]

I'm going to build a coffee heater for a school project.

I know I have to take advantage of the Joule effect (Joule heating). I am restricted to using a resistance between 330 Ω and 1 MΩ. The voltage supply is 5 V, via USB connection.

If I use resistors, then the circuit won't get hot. If I use diodes, their temperature will increase but not enough.

How do I generate enough heat in the circuit to heat coffee?

closed as unclear what you're asking by Tom Carpenter, Voltage Spike, duskwuff, Brian Carlton, Enric BlancoJun 17 '17 at 11:38

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• Nichrome wire + Electricity. – Tom Carpenter Jun 16 '17 at 22:02
• Probably because you haven't done any research on what the value of the resistance should be $P = I^2R$ – Voltage Spike Jun 16 '17 at 22:22
• What do you mean by "if I use resistors do not heat"? You need to go into more detail about what you've tried. – duskwuff Jun 16 '17 at 22:22
• I need to get enough temperature to heat a coffee – Diego Jun 16 '17 at 22:30
• What value resistors did you try? What was the power supply? What was the power dissipation in the resistors? – Peter Bennett Jun 16 '17 at 22:42

What you are doing is not going to work. Sorry, but there it is.

A USB power supply will put out a maximum (typically) of 1.5 amps. Total power is volts times amps, so this is a maximum of 7.5 watts.

However, with a 330 ohm resistor, current is 5 volts divided by 330, or .015 amps. Power is then 5 volts time .015 amps, or .075 watts, and this is just not enough to warm anything.

If you were to connect a bunch of resistor is parallel, and all were 330 ohms, each would produce .075 watts separately, so 100 of them would produce the maximum power.

So your demand that 330 ohm to 1 megohm resistors must be used condemns you to failure.

If you study physics you should learn that

$$t = \frac {\Delta T \times m \times SHC}{P}$$

where P is power in watts (W), t is time in seconds (s), SHC is specific heat capacity (kJ/kgK) and m is mass (kg). SHC for water is 4.2 kJ/kgK.

Testing for a 2.1 kW kettle with one litre of water at room temperature and rearranging the formula we get

$$t = \frac {\Delta T \times m \times SHC}{P} = \frac {(100 - 20) \times 1 \times 4.2}{2.1} = 80 \times 2 = 160 \; s$$ So a bit under three minutes to boil a litre of water from room temperature. This sounds about right.

Since you want to know what power to use you should rearrange the formula:

$$P = \frac {\Delta T \times m \times SHC}{t}$$

Plug in your $\Delta T, \; m, SHC$ and $t$ and calculate the power you think you need.

If you know your supply voltage you can now work out the resistor value from

$$R = \frac {V^2}{P}$$

Finally you can work out the current drawn from your battery from

$$I = \frac {P}{V}$$

All of the above assumes a well insulated container. An open top or conducting sides will cause heat loss.

The power supply is with a USB cable.

You can probably get 1 A at 5 V from a USB power supply. From $P = VI$ you can see that you have 5 W for your heater. Plug that into the formula for calculating $t$ and see if you think heating coffee from a USB port is feasible.

It sounds like you're using the wrong resistors. The power, P is given by

P = V²/R

or

P = I²R

Use whichever is most useful to you. Swap the equations around using basic algebra to work out whichever term you don't know (probably the resistance you need).

Depending on how much coffee you're wanting to heat, you might need anything from a few tens of watts up to a few kilowatts.

You don't say what your power source is. Wiring a home-made coffee heater directly to the mains is probably a bit dangerous, so a bench power supply might be more sensible.

If you're using off-the-shelf resistors, make sure they are rated for the power you intend to produce, with a bit of safety margin on top.

Consider the Max current on the power supply and use maybe 100 watts with thermal pasted and clamped ring heater insulated nichrome wire Maybe 5A at 12V will keep it warm, but not boil cold water.

Try 50 Watts from 5V 50 Watt supply or more (ATX PSU is good)

using V^2/W=R = 25/50 = 0.5 Ohms net resistance = 330 ohms / 660 resistors in parallel = 0.5 Ohm . This means 1/4W resistors with < 1/10W per R but since touching side to side over the whole plate it should get to 85'C . For safety add a thermistor or wire 100 resistors at a time and try to fit them on a baseplate in parallel strings so you can always switch off the outer strings sepearately.

like this and use 24 to 30 AWG magnet wire also embedded in the epoxy.

What I'd do is figure how to solder 100 resistors together like firecrackers so that the ends do not short on the plate. using the cut resistor wires to make a bar across as many as it will reach in a neat flat bundle like firecrackers. In bulk there are only about a penny each and 10% of that in high volume.

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