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Boiling water via usb port

Idontcareforpie

Junior Member
May 15, 2009
12
0
0
I got into an argument with a business major yesterday about a product that could boil water via a usb port. Below is the math I did but at the end is a product that seems to violate the formulas. Did I make a mistake? (other than arguing a science topic with a non science person)

g * (temp change needed °c) * (specific heat of water 4.186J/g °C) = J


10oz of water (1 fluid oz = 29.6ml && 1ml water = 1g) * 29.6g = 296g water


296g * (100c-30c) * 4.186J/g °C = 86733.9 J

USB port = 5v at a max of .5amp


w = VI so USB port max power is 5*.5 or 2.5watts


J = ws

s = J/w

86733.9/2.5w = 34693s = 9.64 hours

This is in line with what I thought it would be but then I found this online



http://www.vat19.com/dvds/usb-warmer-cooler.cfmhttp://www.vat19.com/dvds/usb-warmer-cooler.cfm

Which claims to heat up to 140f. Doing the math again with that I come up with just over 2 hours to heat it.

So what am I missing? The product isn't just bogus is it?

edit: I also am assuming ideal conditions. I don't think the race is close enough to bother with the added hassle.
 

Idontcareforpie

Junior Member
May 15, 2009
12
0
0
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.
 

2Xtreme21

Diamond Member
Jun 13, 2004
7,045
0
0
Your math does seem sound... although this is HIGHLY theoretical. The outside temperature of the room is going to have more of an impact on the temperature of the water, than anything produced from the port.
 

Zenmervolt

Elite member
Oct 22, 2000
24,520
5
81
Originally posted by: Idontcareforpie
I got into an argument with a business major yesterday about a product that could boil water via a usb port.
The only question I have is this: At what rate does the water dissipate the heat energy? If, at any temperature between ambient and boiling, the water is radiating at least 2.5 watts of energy, then it will not boil.

Remember that, as the water heats up it begins to constantly lose energy by radiating it to the surroundings.

Also, your assumptions assume a standard resistance heating element, which the device in question does not have. The device is most likely a Peltier-effect heat pump, which does not create heat energy, it merely transfers heat energy from one area to another. Because of this, it is theoretically possible for a heat pump that draws 2.5 watts of current to transfer more than 2.5 watts of heat to its element (in ideal conditions), so, depending on the quality of the item, 2.5 watts may be a low estimate of its heating ability.

ZV
 

Idontcareforpie

Junior Member
May 15, 2009
12
0
0
Originally posted by: Zenmervolt
Originally posted by: Idontcareforpie
I got into an argument with a business major yesterday about a product that could boil water via a usb port.
The only question I have is this: At what rate does the water dissipate the heat energy? If, at any temperature between ambient and boiling, the water is radiating at least 2.5 watts of energy, then it will not boil.

Remember that, as the water heats up it begins to constantly lose energy by radiating it to the surroundings.

Also, your assumptions assume a standard resistance heating element, which the device in question does not have. The device is most likely a Peltier-effect heat pump, which does not create heat energy, it merely transfers heat energy from one area to another. Because of this, it is theoretically possible for a heat pump that draws 2.5 watts of current to transfer more than 2.5 watts of heat to its element (in ideal conditions), so, depending on the quality of the item, 2.5 watts may be a low estimate of its heating ability.

ZV
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.

Seems to me using the exhaust from the power supply would be more useful.

Do you think there would be any way to approach boiling using a device that is just plugged into a usb port?

Thanks for all the replies, I had hoped to learn something today myself as well!
 

octopus41092

Golden Member
Feb 23, 2008
1,839
0
76
That would be at 100% efficiency and only if the water you're heating up is in a completely isolated environment. Realistically the heat would transfer from the water to the cup to the air and from the water to the air. So.... I don't think it would actually be possible or it would take a hell of a longer time than just 9.64hrs
 

So

Lifer
Jul 2, 2001
25,915
2
81
Originally posted by: Zenmervolt
Originally posted by: Idontcareforpie
I got into an argument with a business major yesterday about a product that could boil water via a usb port.
The only question I have is this: At what rate does the water dissipate the heat energy? If, at any temperature between ambient and boiling, the water is radiating at least 2.5 watts of energy, then it will not boil.

Remember that, as the water heats up it begins to constantly lose energy by radiating it to the surroundings.

Also, your assumptions assume a standard resistance heating element, which the device in question does not have. The device is most likely a Peltier-effect heat pump, which does not create heat energy, it merely transfers heat energy from one area to another. Because of this, it is theoretically possible for a heat pump that draws 2.5 watts of current to transfer more than 2.5 watts of heat to its element (in ideal conditions), so, depending on the quality of the item, 2.5 watts may be a low estimate of its heating ability.

ZV
I've never taken a class that discussed peltiers, but this smacks of perpetual motion to me. Put 2.5W of energy into a peltier, get >2.5W out. If that were the case, why aren't we using it to solve the energy crisis?
 

dullard

Elite Member
May 21, 2001
22,345
701
126
Originally posted by: Idontcareforpie
86733.9/2.5w = 34693s = 9.64 hours
...
Which claims to heat up to 140f. Doing the math again with that I come up with just over 2 hours to heat it.
Your 9.6 hours is correct if the water couldn't lose heat to the environment. In reality, it'll take much longer (if at all possible).

As for the 140°F temperature, that is the temperature of the USB pad, not the temperature of the water. The water would be significantly colder.

Yes, it is possible to boil water with a USB port power only. It just isn't possible to boil a useful (for drinking) amount of water. If you try your math again with 0.1 ounces, you'll see it can happen quite quickly and easilly. It begs the question though as to WHY you'd want to boil such a small amount of water.
Originally posted by: Zenmervolt
Also, your assumptions assume a standard resistance heating element, which the device in question does not have. The device is most likely a Peltier-effect heat pump, which does not create heat energy, it merely transfers heat energy from one area to another. Because of this, it is theoretically possible for a heat pump that draws 2.5 watts of current to transfer more than 2.5 watts of heat to its element (in ideal conditions), so, depending on the quality of the item, 2.5 watts may be a low estimate of its heating ability.
Peltier devices pump energy temporarilly but they quickly reach equilibrium where they no longer pump energy but they act SOLELY as a resistive heater. Temporarilly it may be a couple more watts (but that will last only a second or two). Soon the peltier device will be just like a 2.5W heater.

The only way for them to continuously pump energy is if you had another device that helped transfer heat (such as a fan if you wanted to cool the water or another heater if you wanted to heat the water). The external device for heating water would need more than 2.5W of energy if you wanted it to do any significant work as well. So you come back to square one with the peltier acting solely as a resistive heater.
 

Zenmervolt

Elite member
Oct 22, 2000
24,520
5
81
Originally posted by: dullard
Originally posted by: Idontcareforpie
86733.9/2.5w = 34693s = 9.64 hours
...
Which claims to heat up to 140f. Doing the math again with that I come up with just over 2 hours to heat it.
Your 9.6 hours is correct if the water couldn't lose heat to the environment. In reality, it'll take much longer (if at all possible).

As for the 140°F temperature, that is the temperature of the USB pad, not the temperature of the water. The water would be significantly colder.

Yes, it is possible to boil water with a USB port power only. It just isn't possible to boil a useful (for drinking) amount of water. If you try your math again with 0.1 ounces, you'll see it can happen quite quickly and easilly. It begs the question though as to WHY you'd want to boil such a small amount of water.
Originally posted by: Zenmervolt
Also, your assumptions assume a standard resistance heating element, which the device in question does not have. The device is most likely a Peltier-effect heat pump, which does not create heat energy, it merely transfers heat energy from one area to another. Because of this, it is theoretically possible for a heat pump that draws 2.5 watts of current to transfer more than 2.5 watts of heat to its element (in ideal conditions), so, depending on the quality of the item, 2.5 watts may be a low estimate of its heating ability.
Peltier devices pump energy temporarilly but they quickly reach equilibrium where they no longer pump energy but they act SOLELY as a resistive heater. Temporarilly it may be a couple more watts (but that will last only a second or two). Soon the peltier device will be just like a 2.5W heater.

The only way for them to continuously pump energy is if you had another device that helped transfer heat (such as a fan if you wanted to cool the water or another heater if you wanted to heat the water). The external device for heating water would need more than 2.5W of energy if you wanted it to do any significant work as well. So you come back to square one with the peltier acting solely as a resistive heater.
Interesting. I did not know this and thought it worked similar to an evaporative heat pump.

ZV
 

Gibson486

Lifer
Aug 9, 2000
18,379
1
0
Originally posted by: So
Originally posted by: Zenmervolt
Originally posted by: Idontcareforpie
I got into an argument with a business major yesterday about a product that could boil water via a usb port.
The only question I have is this: At what rate does the water dissipate the heat energy? If, at any temperature between ambient and boiling, the water is radiating at least 2.5 watts of energy, then it will not boil.

Remember that, as the water heats up it begins to constantly lose energy by radiating it to the surroundings.

Also, your assumptions assume a standard resistance heating element, which the device in question does not have. The device is most likely a Peltier-effect heat pump, which does not create heat energy, it merely transfers heat energy from one area to another. Because of this, it is theoretically possible for a heat pump that draws 2.5 watts of current to transfer more than 2.5 watts of heat to its element (in ideal conditions), so, depending on the quality of the item, 2.5 watts may be a low estimate of its heating ability.

ZV
I've never taken a class that discussed peltiers, but this smacks of perpetual motion to me. Put 2.5W of energy into a peltier, get >2.5W out. If that were the case, why aren't we using it to solve the energy crisis?
You can't take more energy than you put in. Law of convservation of energy and thermodynamics says so.

 

Zenmervolt

Elite member
Oct 22, 2000
24,520
5
81
Originally posted by: So
I've never taken a class that discussed peltiers, but this smacks of perpetual motion to me. Put 2.5W of energy into a peltier, get >2.5W out. If that were the case, why aren't we using it to solve the energy crisis?
I mistook a peltier as a solid state equivalent to a traditional heat pump, see dullard's post below.

However, a traditional heat pump can transfer more heat than the energy used to run the fans and compressor. However, remember that this is transferred heat energy, not created heat energy. A standard air conditioner is effectively a "one-way" heat pump (a true heat pump can both cool and heat); it absorbs heat energy at the evaporator and transfers that heat to the condenser. Heat that is pulled out of the air around the evaporator is radiated by the condenser and it is possible that this transferred heat is greater in quantity than the total energy input required to run the compressor and the fans.

This is not perpetual motion because the "excess" heat energy is not being "created" out of nowhere, but simply transferred from one medium to another. For a heat pump to work, one area must be cooled to an equal degree that the other area is heated.

ZV
 

bobsmith1492

Diamond Member
Feb 21, 2004
3,880
2
81
Originally posted by: So
Originally posted by: Zenmervolt
Originally posted by: Idontcareforpie
I got into an argument with a business major yesterday about a product that could boil water via a usb port.
ZV
I've never taken a class that discussed peltiers, but this smacks of perpetual motion to me. Put 2.5W of energy into a peltier, get >2.5W out. If that were the case, why aren't we using it to solve the energy crisis?
No, in a peltier, you put in say 10W of heat to transfer 1W out of the cold side, leaving 11W exiting the hot side. It just moves thermal energy from one side to the other; they're maybe 10% efficient.
 

Jeff7

Lifer
Jan 4, 2001
41,602
10
81
You could boil water with an LED if you wanted to. A good one might consume 70mW.

You'd just need an insanely good insulating material to prevent any energy from leaving the water, and a lot of patience. ;)

 

silverpig

Lifer
Jul 29, 2001
27,708
3
76
Originally posted by: bobsmith1492
Originally posted by: So
Originally posted by: Zenmervolt
Originally posted by: Idontcareforpie
I got into an argument with a business major yesterday about a product that could boil water via a usb port.
ZV
I've never taken a class that discussed peltiers, but this smacks of perpetual motion to me. Put 2.5W of energy into a peltier, get >2.5W out. If that were the case, why aren't we using it to solve the energy crisis?
No, in a peltier, you put in say 10W of heat to transfer 1W out of the cold side, leaving 11W exiting the hot side. It just moves thermal energy from one side to the other; they're maybe 10% efficient.
And if you don't dissipate that 11W of heat it'll heat up both the hot and cold sides, and you'll end up with two hot sides and no heat transfer.

OP: The largest loss will come from evaporation. You'll heat up the water to maybe 32 C initially but the latent heat of evaporation will cool the water down as you heat it up. There'll be some sort of equilibrium reached and you'll evaporate the water slowly over say 6 hours.
 

halik

Lifer
Oct 10, 2000
25,708
1
0
I'm pretty sure the heat transferred from the resistor coil to the USB wiring and connectors would melt all the plastic bits in sight.
 

Paperdoc

Golden Member
Aug 17, 2006
1,833
90
91
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".
 

eplebnista

Lifer
Dec 3, 2001
24,150
35
91
Originally posted by: Jeff7
You could boil water with an LED if you wanted to. A good one might consume 70mW.

You'd just need an insanely good insulating material to prevent any energy from leaving the water, and a lot of patience. ;)
Text :p
 

Eli

Super Moderator | Elite Member
Super Moderator
Oct 9, 1999
50,425
5
81
Originally posted by: Paperdoc
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".

:thumbsup:
 

Sea Moose

Diamond Member
May 12, 2009
6,936
6
76
The kettle in my kitchen is 2100watts it takes about 5 or 6 minutes to boil water.
that device would take hours to boil water.
I think it just helps keep it warm.

I have the enthalpy chart for water somewhere... trying to find it
 
Sep 29, 2004
18,651
64
91
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.
 

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