Boiling water via usb port

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Sea Moose

Diamond Member
May 12, 2009
6,936
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Originally posted by: IHateMyJob2004
Originally posted by: Idontcareforpie
Originally posted by: erub
100% efficiencies, horray! *ahem*
Oh yeah, I forgot to mention that. I don't really feel like trying to take anything like that into account. Assuming ideal situations I think will be enough to prove my point.

Off to update original.
Well, with the perfect insulator, I could stick a cup up my arse and make the water boil.
pic
 

Red Squirrel

No Lifer
May 24, 2003
56,230
7,586
126
www.uovalor.com
You need a flux capacitor and rectifier in the circuit. What you do is (through modern electronics) produce a sine wave, then as the sine wave goes down, go back in time (only a time warp local to the device) to when the sine wave was at it's peak. So what you are doing is amplifying the voltage in micro time warps. So you have the result of 100 usb ports within a single "tick" of time. By further tweaking it you can even get the power of 1000 usb ports as the energy it produce 1 clock tick ago is actually being produced now, and then next clock tick, that energy that was produced (double) is being produced at the same time as now, so now you have 3x, and so on...

You might be able to find schematics online to build such circuitry. Be careful on the timer circuit that stops the time warping at every 1/100th of a second or you can end up with very disastrous results.

Oh and make sure to wait a few hours after you built the device before running it as if your timing is wrong it may warp to before it was built, then you have to restart the whole project.


In all seriousness, the device probably assumes the water is already boiling when it is applied to it. :p
 

Erithan13

Senior member
Oct 25, 2015
218
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Mod edit: thread was bumped by a spammer. Spam removed. -Admin DrPizza
Well, with the perfect insulator, I could stick a cup up my arse and make the water boil.
If it's a perfect insulator how does the heat get in? And even if it could it'd never get warmer than the surroundings.

(proof left as an exercise to the reader. Cup, arse and insulator provided at your own expense).
 
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CZroe

Lifer
Jun 24, 2001
23,637
633
126
Many USB devices like USB-power HDDs, disc drives, fans, and cup warmers operate way outside of specifications and draw much more than 0.5 Amp. Yes, 500mA is the standard specification, but it certainly is not the maximum such a device might draw.

Also, even .5A can boil water... if you use use that to trickle-charge a big battery!
 
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Red Squirrel

No Lifer
May 24, 2003
56,230
7,586
126
www.uovalor.com
I guess if you can get an element to at least 100 degrees with 500ma you can boil water, it's just a matter of how long it will take to boil a reasonable volume. It also needs to bring the temp higher than it takes to dissipate.
 

master_shake_

Diamond Member
May 22, 2012
6,430
291
121
What are you talking about?
start with a small amount of water and get it to boil.

then add a smaller amount of water than you started with and bring it to a boil.

continue until desired.

wouldn't that make a difference?
 

who?

Platinum Member
Sep 1, 2012
2,327
42
91
Hmmm, so you think the coffee heater is using the air in the room to heat the beverage up? Can a heat pump heat beyond the temp of the room.

Energy involves mass as well as temperature. If the heat pump can concentrate the energy it will raise the temperature higher. Heat pumps in houses can keep the interior at 70 degrees F even when pumping heat from the colder outdoors.
 

gorcorps

aka Brandon
Jul 18, 2004
30,642
376
126
Did anybody actually read the full product description? It's pretty damn funny
 

CZroe

Lifer
Jun 24, 2001
23,637
633
126
start with a small amount of water and get it to boil.

then add a smaller amount of water than you started with and bring it to a boil.

continue until desired.

wouldn't that make a difference?

Not at all. Think about it.
 
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Bjokke

Junior Member
Jun 1, 2020
3
0
6
Your calcs are OK except for the second case. The claim is up to 140F, which translates to 60C. You started by assuming the initial temp is 30C so that you had to heat by 70C. But if your target is 60C, not 100, then the temp rise is 30C, not 70, and the heating time calc comes to 4.13 hours (not just over 2).

Hugest problem is your calcs is heat loss from the heating cup. You hypothesize "ideal case" by which you appear to include NO heat loss by perfect insulation. Ain't gonna happen, ever! Even with good insulation you would surely reach a temperature of the mug and coffee that creates a heat loss of 2.5 watts exactly offsetting the heat input, and temp would never rise further. Moreover, because of this real heat loss rate (which depends on the cup temp), the time to get up to that equilibrium temp would be MUCH longer than your 9 hour estimate. By the way, my suspicion is that, on the website you linked for the commercial heating pad, what they did not tell you is that the device may be able to KEEP at mug of coffee at 140F, but only if it is put on the heating pad already at that temperature. In other words, they estimate that the heating input rate their unit provides will just manage to match the heat loss rate from the mug at 120 to 140F.

Another big problem with your calcs is you ignored the mug. To heat the 296 g of water (coffee, whatever), you also will have to heat the mug itself, unless you can produce a perfect mug that never steals heat from its contents. The mass of the mug probably is similar to the mass of the contents. If we knew the heat capacity of the mug material, we could calculate its heat requirements to keep up with the warming coffee, and that will be in the same ballpark. So now your time estimate is roughly doubled, but not known precisely.

By the way, you also made a slight error in stating the problem. You said, "Boiling water via USB port". What you actually calculated was how to get the liquid water temperature from 30C to 100C. That is, the water is as hot as boiling water, but it has not actually been boiled to convert it to water vapor. And for purposes of drinking hot coffee, that is just fine. But to BOIL water means to finish the conversion of it to water vapor, and that's a huge surprise to many people. The process of converting liquid water at 100C to water vapor still at 100C requires more heat input for the change of state. The process is called vaporization (no surprise there!), and the Heat of Vaporization of water is 100 times the Heat Capacity of water! It's a strange coincidence, actually. With a Heat capacity of 4.186 J per gram-degree, it takes 418.6 Joules to raise the temperature of one gram of liquid water from its freeing point, 0C, to its boiling point, 100C. But then to convert that same gram of water liquid to vapor takes an additional 418.6 Joules of energy!

Oh, and the comments about 100% efficiency are a little misleading. Efficiency usually is the word to describe how much of the energy input (electrical from the USB port) actually gets converted into usable heat. IF the heating device you postulate is a simple resistor, it is guaranteed to be 100% efficient. ALL of the energy flowing through a resistor is converted to heat, and none gets lost, destroyed, or stored. On the other hand, once that conversion to heat is done, all the other processes concerning where the heat goes and how much gets lost from the coffee to the surroundings have a big impact, as all have said. If you include those processes in the concept of "Efficiency" of the system, then it certainly is not "100% Efficient".
So what if you change the vessel. Lets say a vacuum insulated stainless steel waterbottle, and use a usb powered heating-coil built into a screwcap. Closing of the vessel and trapping the heat. Perhaps even, two coils at once, powered by two usb ports on the same battery. Could you then heat 350ml of water (12C) to +70C ?
 

Paperdoc

Golden Member
Aug 17, 2006
1,833
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Hypothetically, yes. IF you assume you can actually create a containment vessel for the water that will never allow any of the heat inside to escape (and a vacuum-insulated bottle like the old "Thermos" glass units is pretty good, although not perfect) and if you can manage to add wiring connections through the bottle to power the heater that also do not allow any heat escape, then this device you create is a 100% efficient heating system. That is, NO heat escapes, and all of the energy added via electrical heat is trapped inside in the water and the containment vessel itself. Note that, in addition to the water (of whatever mass you decide), all of the heating element and associated wiring will need to be heated, too, as well as at least the interior portion of that insulated containment vessel. So the total amount of heat to be added is more than the water's requirement alone, thus increasing the time required. But in principle this is possible in the ideal situation.

Of course, we will NOT achieve ideal. Even a very good insulated containment vessel will have a modest rate of heat loss through its walls. It may not be as much as the heater capacity (2.5W), but even if it is only 0.5 W, that reduces your net heating rate by 25% and this increases the time, again, by big factor. In my student days in Physical Chemistry we used Calorimeter equipment to measure heat generation and loss. The equipment was pretty good at trapping heat, but not perfect. One required step in such use was to do a Calorimeter calibration run in which we actually measured the rate of heat loss to a room of constant known temperature from an interior at various temperatures. Then we could make any necessary correction calculations is processing experimental results.

I'll give you a couple examples of real-world circumstances in this scenario. This refers to my experience with heated coffee mugs for auto travel. Their basic structure is a double-wall mug with a small electrical heating coil mounted inside to keep your coffee warm as you drive. The heater is powered by a cord to the car's cigarette lighter socket. The mug is NOT well insulated because it can get sufficient power from the car electrical system. In my experience, such a device with a heater rated for 2.0 W (12 VDC at 1.67A) is not suffucuent - the coffee that you put into that jug quite hot will cool down to luke-warm in an hour. However, if you get a model with a heater of 2.5 to 3.0 W, it CAN keep the coffee very hot for many hours. The best I've used had a heater of over 3.5 W capacity and a small thermostat to turn off the heater when the coffee gets too hot. That one actuall could heat up coffee that started out too cool. So, from experience with low-cost coffee containers and heaters that are FAR from ideal perfect insulators (and hence cheap enough to sell to any normal person) you need at least a 3W heater. That's close to what a USB2 port can provide. What that CAN achieve is, IF you put HOT coffee in it, at constant full power it will probably keep your coffee pleasantly warm enough to drink for a long time.
 

CZroe

Lifer
Jun 24, 2001
23,637
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126
If you could contain the heat anywhere near 100% then you have to stop evaporation from carrying away heat. That means it would need to be sealed and be under pressure but... surprise surprise: pressure raises the boiling point. You would have to have a pressure vessel that retains nearly 100% of the heat and then suddenly open it to get any boil at all. Even then, the energy that boils off will exceed the input and stop boiling very quickly... almost instantly unless it is a very small volume that would boil away completely in no time. I dont think anyone wants to boil less than a thimble-full. ;)
 

Arx Allemand

Member
Sep 24, 2019
45
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The power output of a USB port can certainly boil water!
Not by resistance heating or using a thermoelectric pile...

However, if you put the water in a sealed vessel and used a small pump that ran on 5V 1A power, it could evacuate the air to the point where the water would boil furiously...at room temperature! ;)
 
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Bjokke

Junior Member
Jun 1, 2020
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Ok thanks all! One more try. :) new tech, new bottle. What if you combine an 18w output usb C, with a 500ml insulated growler. See picture. Could you then heat 350ml of water (12C) to be 100C ? After testing the growler i can say it holds heat fairly well. 95c water was still 60c after 12h. And it hold pressure well. Perhaps a pressure valve is needed like in a kitchen-pressurecooker.


1593847612775.jpeg
 

Paperdoc

Golden Member
Aug 17, 2006
1,833
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USB-C may not make 18W available - that is more than a standard USB3 port - but some certainly can because they are designed to exceed USB3 standards.

Now we need to refine exactly what you want to achieve. Step one is raising the temperature of the water from 12C to 100C, and that takes 1 cal per gram per degree C. For your example, 350 g water heated by 88 degrees will require an input of 30,800 cal. with NO loss of heat during the heating process. How much will really be lost from that vessel? Cannot know for sure without more data, but it will be more than zero. 1W = 14.33 cal/min heating rate, so for that total heat you would have to operate at 18 W for 2149 minutes or 1.5 DAYS. Then add on the adjustment for heat loss during that time - guess that will take us to 3 days or more.

THEN we have the question of whether you want just hot water at 100C, or you actually want to BOIL some water and convert some of it to steam. If you are trying to make some instant coffee, you do not need to actually BOIL any water. So far, just heating the water over that temp range needed 88 cal/gram. Now, to CONVERT liquid water at 100C to water in gas form (steam) still only at 100C is a change of state, and that requires an additional 100 cal PER GRAM - more than what it took to get the liquid to 100C! It's an interesting exercise in possibility, but points to the futility of trying in real practice.
 

Bjokke

Junior Member
Jun 1, 2020
3
0
6
USB-C may not make 18W available - that is more than a standard USB3 port - but some certainly can because they are designed to exceed USB3 standards.

Now we need to refine exactly what you want to achieve. Step one is raising the temperature of the water from 12C to 100C, and that takes 1 cal per gram per degree C. For your example, 350 g water heated by 88 degrees will require an input of 30,800 cal. with NO loss of heat during the heating process. How much will really be lost from that vessel? Cannot know for sure without more data, but it will be more than zero. 1W = 14.33 cal/min heating rate, so for that total heat you would have to operate at 18 W for 2149 minutes or 1.5 DAYS. Then add on the adjustment for heat loss during that time - guess that will take us to 3 days or more.

THEN we have the question of whether you want just hot water at 100C, or you actually want to BOIL some water and convert some of it to steam. If you are trying to make some instant coffee, you do not need to actually BOIL any water. So far, just heating the water over that temp range needed 88 cal/gram. Now, to CONVERT liquid water at 100C to water in gas form (steam) still only at 100C is a change of state, and that requires an additional 100 cal PER GRAM - more than what it took to get the liquid to 100C! It's an interesting exercise in possibility, but points to the futility of trying in real practice.
Great feedback, thank you! I was hoping one day to replace a pocket gasstove to boil water at camping trips with a powerbank. No open fire, no gas canister, recharged by sun if needed etc. Boiling the water would be needed for purification as well as for food. (Freezedried)
1,5 days at best is still way to long to reach the desired result. Perhaps in a few years ;).
 

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