Can I Heat Water with 5 Volts and 2-3 Grounds?

[abm04786]

Just have a question: how can I make a steady temperature to heat water or liquid where I have 5 volts with 2 or 3 grounds as a source only?
Please explain the things that can be required to increase things such as resistance and current and their source such diodes or batteries with the require volts or any such stuff.

Want to heat water or other liquid. Have only 5 volts and 2 or 3 grounds. Please explain the things how to get the standard volts , current and resistance ratio such as the supply to our homes through 5 volts and 2 or 3 grounds as my source which after assembled and good enough to heat water or liquid.

 

[russ_watters]

Welcome to PF!

We need more information in order to adequately help:

How much water?
How hot?
How cold does it start?
How fast do you want to heat it?
How precisely do you want to control it?
What Is the power source?
How many watts can it produce?
Is the container insulated?

 

[amanno]

You should also make menition of what kind of source the 5V is coming from... a power supply? Battery? That is help determine the power you have available.

 

[abm04786]

Welcome to PF!

We need more information in order to adequately help:

How much water?
How hot?
How cold does it start?
How fast do you want to heat it?
How precisely do you want to control it?
What Is the power source?
How many watts can it produce?
Is the container insulated?

1) Water :- a cup of water means as usual cup that we all use to drink and up to 1/2 liter
2) hot :- nearly about to make it a little warm about 90-95 Celsius and also tell me about boiling or boiling temperature or to get reach to boil of water or liquid
3) cold:- room temperature or stuff kept in normal ac rooms
4) fast:- normal heat as we heat a cup of water or liquid
5) control:- not get boiled to fast normally at a standard home electric supply control.
6)Power source:-It's fixed its a 5 volts in a wire as supply it can't be changed with 2 or 3 grounds but other stuff such diodes or resistors can be added to it to the wire.
7)Watts :-I don't know I have a fixed 5 volt fixed supply in a wire and 2 or 3 grounds. Grounds may vary but 2 or more it only.
8) Its not container its a cup or a vessel that can be made conductive.

One think I have is "use of induction" but should be through only 5 volts and 2 or 3 grounds.

 

[abm04786]

You should also make menition of what kind of source the 5V is coming from... a power supply? Battery? That is help determine the power you have available.

a battery as a power supply a steady power supply through battery of 5 volts.

 

[davenn]

2 or 3 grounds doesn't really make sense in the context of what you are talking about

I haven't come across any 5 V batteries in my time...
To do any significant heating of a half liter container of water you are going to need a respectable PSU capable for supplying probably at least 5Amps or so. Maybe you should look at a 12V car battery and suitable waterproof heating element if the element needs to be immersed in the water

Dave

 

[abm04786]

2 or 3 grounds doesn't really make sense in the context of what you are talking about

I haven't come across any 5 V batteries in my time...
To do any significant heating of a half liter container of water you are going to need a respectable PSU capable for supplying probably at least 5Amps or so. Maybe you should look at a 12V car battery and suitable waterproof heating element if the element needs to be immersed in the water

Dave

More Details:=

Just Think you have 5volts (fixed) being supply in a wire (which you can increase "volts" adding any electronic stuff to it) and 2 or 3 grounds wire in your hand (you can increase too) to a level that you can generate enough "electricity" to "heat" a cup of water "max 1/4 liter" not more then it.

5volts supply is fixed but it can be increase using any electronic stuff and ground may vary but minimum 2 is required atleast.

This means initial are fixed that is there is 5volts supply in a wire and 2 grounds.You have 5 volts and 2 grounds you can added any electonics stuff to increase volts and grounds to make it enough electronic to generate electricity to heat maximum 1/4 liter of water to a temperature of about 90-95 degree Celsius but not boil it to 100 degree Celsius.

 

[davenn]

More Details:=

Just Think U have 5volts (fixed) being supply in a wire (which you can increase "volts" adding any electronic stuff to it) and 2 or 3 grounds wire in your hand (u can increase too) to a level that U can generate enough "electricity" to "heat" a cup of water "max 1/4 liter" not more then it.

5volts supply is fixed but it can be increase using any electronic stuff and ground may vary but minimum 2 is required atleast.

This means initial are fixed that is there is 5volts supply in a wire and 2 grounds.U have 5 volts and 2 grounds U can added any electonics stuff to increase volts and grounds to make it enough electronic to generate electricity to heat maximum 1/4 liter of water to a temperature of about 90-95 degree Celsius but not boil it to 100 degree Celsius.

Sorry abm04786

that really doesn't make any sense at all

maybe draw a diagram of what you are planning and see if that helps us all understand your plan :)

Dave

 

[abm04786]

Here is the image with it wires:-
http://postimg.org/image/w72jv2ue9/
Here is the link to image:http://postimg.org/image/w72jv2ue9/
the image tells you all the things I its only the electronic stuff is needed to add to it to increase its required like resistors,diodes etc.

 

[mfb]

Well, one basic problem is power: batteries have some limit for the power they can provide. Without knowledge about those batteries, it is hard to get specific.

Two examples:
If those batteries can provide 1A (and not more), you get a maximum of 5W. Heating 1/2 liter of water from room temperature to 90°C with 5W needs at least 140kJ, or 8 hours with 5W.
In reality, your water will lose heat to the environment - it will never get warm.
If those batteries can provide 10A, you get a maximum of 50W. Heating 1/10 liter of water from room temperature to 90°C with 50W needs 9 minutes. That might be possible.

Your image has no scale, so I don't know if the cable can support 10A (without significant voltage drop inside).

the image tells U all the things

Actually, it tells nearly nothing. You still did not provide details about the power supply.
Please use proper english here, "U" is not a word.

 

[OmCheeto]

Here is the image with it wires:-

Here is the link to image:http://postimg.org/image/w72jv2ue9/
the image tells U all the things I its only the electronic stuff is needed to add to it to increase its required like resistors,diodes etc.

Ah ha! A USB powered coffee warmer.

It looks like the Chinese have invented one:

http://ip.com/patfam/en/44269186

hmmm...

USB 2.0 is rated 0.5 amps @ 5.0 volts = 2.5 watts
Water has a heat capacity of 4.2 Joules/(gram * degrees Kelvin)
A cup of water is 240 grams
°C = K-273
Watt = Joule / Second

Say
T_c = 15 °C
T_h = 90 °C
Δt = 75 °C = 75 K
(note: 75 °C ≠ 75 K, Δt can be read as; "the change in temperature"

I assume we want the time it will take to heat the water, so:

Seconds = Joule/Watt

and

joules = (K * grams)*4.2 = (75 * 240)*4.2 = 75,600 joules

Seconds to heat the cup of water = 75,600 joules/2.5 watts = 30,240 seconds = 504 minutes = 8.4 hours, assuming a perfectly insulated cup.

Ah! I almost forgot the resistor
R = E/I = 5/.5 = 10 Ω

hmmm... Experiment!

75,600 joules / 75 K = 1000 joules / K
1000 joules / K * 0.556 K / °F = 556 joules / °F
5560 joules / 10°F change

required components: 1 coffee cup, 1 cup water, 1 clock, 1 thermometer, 1 mfb, and 1 jim hardy
water heated in microwave to boiling
cup brought back to coffee table for data entry and analysis (~30 seconds)

ambient temp = 68'F

07:42:00 204'F
07:43:30 194'F in __90 s --> 5560/_90 = 61.8 watts
07:45:20 184'F in _110 s --> 5560/110 = 50.5 watts
07:48:00 174'F in _160 s --> 5560/160 = 34.8 watts
07:51:00 164'F in _180 s --> 5560/180 = 30.9 watts
07:55:00 154'F in _240 s --> 5560/240 = 23.2 watts
07:59:00 144'F in _240 s --> 5560/240 = 23.2 watts
08:05:00 134'F in _360 s --> 5560/360 = 15.4 watts
08:11:00 124'F in _360 s --> 5560/360 = 15.4 watts (started a pot of coffee)
08:21:00 114'F in _600 s --> 5560/600 = 9.27 watts (started making edits to this post)
08:33:00 104'F in _720 s --> 5560/720 = 7.72 watts (replaced Kelvin temps with 5560 constant)
08:52:00 _94'F in 1140 s --> 5560/1140 = 4.88 watts (put in comment about pitch dripping)
09:17:00 _84'F in 1500 s --> 5560/1500 = 3.71 watts (made some minor edits)
09:30:00 _80'F in _780 s ------------> = 2.85 watts (decided the experiment was complete)
11:52:00 _69'F in 8520 s ------------> = 0.71 watts (curses at self!)

It looks as though my system would not even reach 80°F with a 2.5 watt heater.
Ixnay on the USB powered coffee warmer.



[edit: complete rewrite thanks to mfb and jim hardy keeping an eye on me]



"Use of the degree symbol to refer to temperatures measured in kelvins (symbol: K) was abolished in 1967 by the 13th General Conference on Weights and Measures (CGPM)."

 

[mfb]

@OmCheeto: You have to multiply by 4.2 instead of dividing through it. Therefore, your power values have to be multiplied with ~17.5 and you cannot even reach 82F ~ 28°C with 5W.

 

[jim hardy]

Water has a heat capacity of 4.2 Joules/(gram * degrees Kelvin)...
A cup of water is 240 grams...

components: 1 coffee cup, 1 cup water, 1 clock, 1 thermometer...

the water in this experiment requires 57 joules per degree K change...

uhhhh I'm confused somehow - do we multiply or divide?

240 grams X 4.3 joules/gram-degree = 1008 joules/degree K, 560 per degree F ?

Or did you use a tiny cup?

lessee here $\frac{57j/degK}{4.3j/gram-degK}$ = only 13 grams of water? A half ounce?


anyhow if you used a 240 gram cup, a loss of tenF in 90 seconds gives me a higher evaporation number
$\frac{5600}{90} = 62.2$
Your first interval of 5600 joules in 90 seconds = 62 watts, and from your readings two watts would settle at maybe ten deg above ambient...?

But I'm getting awful rusty... one of us may be off by 4.3 squared...

old jim

 

[OmCheeto]

uhhhh I'm confused somehow - do we multiply or divide?

240 grams X 4.3 joules/gram-degree = 1008 joules/degree K, 560 per degree F ?

Or did you use a tiny cup?

lessee here $\frac{57j/degK}{4.3j/gram-degK}$ = only 13 grams of water? A half ounce?


anyhow if you used a 240 gram cup, a loss of tenF in 90 seconds gives me a higher evaporation number
$\frac{5600}{90} = 62.2$
Your first interval of 5600 joules in 90 seconds = 62 watts, and from your readings two watts would settle at maybe ten deg above ambient...?

But I'm getting awful rusty... one of us may be off by 4.3 squared...

old jim

@OmCheeto: You have to multiply by 4.2 instead of dividing through it. Therefore, your power values have to be multiplied with ~17.5 and you cannot even reach 82F ~ 28°C with 5W.

Fixed! I think.

 

[Integral]

;

abm04786, is Om correct? Are you talking about using USB as your power source? If so, why have you not said that?

Om has shown that using a computers USB port you cannot make the heater you want. If the other end of that cable you linked to is NOT a USB connector then you need to tell us EXACTLY what it is connected to. The voltage really does not matter much, what is critical is the current, in Amps that your device produces. Without that knowledge no one can help you.

When I say "voltage does not matter" I mean it can be 5V, 12V or whatever, as long as we KNOW what it is.

 

[jim hardy]

sorry - mfb posted while I was still ciphering and worrying...

glad it's ok now...

and thanks for the kind words !

old jim