# Thread: boil water from room temp using regular batteries?

1. Hello everyone,

I want to warm up/boil water using batteries. I got the kJ it takes to boil water and i am stuck at how to find the kJ each battery has if i have the voltage and the capacity (mAh) given?

Any help will be appreciated!

2.

3. Originally Posted by inventor2014
Hello everyone,

I want to warm up/boil water using batteries. I got the kJ it takes to boil water and i am stuck at how to find the kJ each battery has if i have the voltage and the capacity (mAh) given?

Any help will be appreciated!
Sounds like homework. The clue is: W = V x i. That should be enough…..

4. Originally Posted by exchemist
Originally Posted by inventor2014
Hello everyone,

I want to warm up/boil water using batteries. I got the kJ it takes to boil water and i am stuck at how to find the kJ each battery has if i have the voltage and the capacity (mAh) given?

Any help will be appreciated!
Sounds like homework. The clue is: W = V x i. That should be enough…..
there is no homework as semester is out.
and i knew that formula just wanted to make sure the numbers i got made sense
because if my battery is:
3.4 Ah, 3.7V - if i run it for 5min (300s) straight it would give out 40.8 A and that results in 45.3 kJ.

So if water requires 167.4kJ, i will need about 4 of these batteries to accomplish 80C temp increase? that sounds wrong to me

5. Originally Posted by inventor2014
3.4 Ah, 3.7V - if i run it for 5min (300s) straight it would give out 40.8 A and that results in 45.3 kJ.
1) No, the 4.4ah is a low C rate, not a 5 minute rate. You'll only get about 50% of the battery's capacity at those rates.
2) The total joules (watt-seconds) available from the battery at a slow rate is (amps x volts x time) or 12.58*3600=45288 joules, so your above number is correct for a slow discharge. For a fast discharge you'd be around 25 kJ.

So if water requires 167.4kJ, i will need about 4 of these batteries to accomplish 80C temp increase? that sounds wrong to me
Depends upon the amount of water of course but that sounds in the ballpark. Heating water takes a lot of energy.

6. 2) The total joules (watt-seconds) available from the battery at a slow rate is (amps x volts x time) or 12.58*3600=45288 joules, so your above number is correct for a slow discharge. For a fast discharge you'd be around 25 kJ.
okay that make sense! i was calculating per hour and counted as if it's 5min so that's were i messed up.
i was trying to figure out how to get the fast discharge value by including the 5 minutes in the formula but not sure where to put it..
this is wrong obviously: (3.7 v) * (3.4 Ah) * (300 s) = 3.8 kJ

7. Originally Posted by inventor2014
Originally Posted by exchemist
Originally Posted by inventor2014
Hello everyone,

I want to warm up/boil water using batteries. I got the kJ it takes to boil water and i am stuck at how to find the kJ each battery has if i have the voltage and the capacity (mAh) given?

Any help will be appreciated!
Sounds like homework. The clue is: W = V x i. That should be enough…..
there is no homework as semester is out.
and i knew that formula just wanted to make sure the numbers i got made sense
because if my battery is:
3.4 Ah, 3.7V - if i run it for 5min (300s) straight it would give out 40.8 A and that results in 45.3 kJ.

So if water requires 167.4kJ, i will need about 4 of these batteries to accomplish 80C temp increase? that sounds wrong to me
I thought you originally said mAh, didn't you? Has a factor of 1000 gone missing somewhere?

8. Originally Posted by exchemist
Originally Posted by inventor2014
Originally Posted by exchemist
Originally Posted by inventor2014
Hello everyone,

I want to warm up/boil water using batteries. I got the kJ it takes to boil water and i am stuck at how to find the kJ each battery has if i have the voltage and the capacity (mAh) given?

Any help will be appreciated!
Sounds like homework. The clue is: W = V x i. That should be enough…..
there is no homework as semester is out.
and i knew that formula just wanted to make sure the numbers i got made sense
because if my battery is:
3.4 Ah, 3.7V - if i run it for 5min (300s) straight it would give out 40.8 A and that results in 45.3 kJ.

So if water requires 167.4kJ, i will need about 4 of these batteries to accomplish 80C temp increase? that sounds wrong to me
I thought you originally said mAh, didn't you? Has a factor of 1000 gone missing somewhere?
i feel like im confusing myself, i thought 3400 mAh is 3.4 Ah right?
so that's 3.4 A per hour, now if i want to know how many amps per 5 min i just multiply 3.4 by 300 seconds?

9. I can see a flaw here straight away. If you have 3.4 A per hour you need to divide by 12 to get the Amps in five minutes. (12 x 5 = 60 mins), you would only multiply by 300s if your original number was in A per second not A per hour.

10. Originally Posted by inventor2014
i feel like im confusing myself, i thought 3400 mAh is 3.4 Ah right?
so that's 3.4 A per hour, now if i want to know how many amps per 5 min i just multiply 3.4 by 300 seconds?
If you can get 1 A out of a battery for 1.00 hour, you can't get 10 A out of it for 0.10 hours, nor 100 A for 0.01 hours, etc.

To begin with, these battery "capacities" are at some rated amperage (say 100 mA) and perhaps even a duty cycle (maybe 1 h on and 4 h off). The reason is the battery's efficiency drops with increased output and/or duty cycle, otherwise you should be able to get, say, a gazillion horsepower out of a battery for, say, 1 microsecond. If you tried this, the descriptive word would be "cataclysmic", which, even on a small scale, could mean a trip to the hospital or morgue.

For one thing, there's limitations to the electrochemical processes involved.

Then there's something called Peukert's law that probably applies as, if I recall properly, it models some of the inefficiencies of the electrochemical processes (see above).

Also consider the internal resistance of batteries. The battery's internal power loss equals I˛R. As you lower the load resistance in order to increase the current, the internal power consumption increases with the square of the current. As a simple demonstration, lowering the load resistance to the battery's internal resistance means that the battery's power production is split evenly among the resistances, so you're only able to use 50% of what's produced. Unless you want to use the hot/exploding battery as a source of heat. HINT: Don't do it.

Batteries are weird devices in several ways. They can also kinda bounce back if given time to "rest", which is why duty cycle is a consideration. You might operate a radio for 20 hours (straight) on fresh batteries before it "dies". If you operate it for 1 hour, then "rest" it for 4 hours (ie, 20% duty cycle), it can operate longer, say, 25 hours of combined run time before it "dies"— of course, over a total period of about 120 hours.

Long story short, of all the forms of energy sensed by humans (sonic, visual, etc), thermal energy requires the most energy to produce noticeable effects, which is why consumers complain about their heating bills, and why parents yell at their kids for taking long, hot showers. Currently (no pun intended), batteries are so inefficient that to boil a cup (8 fl oz, 250 mL) of water requires, what would seem to us to be, a ridiculous amount of batteries.

11. Originally Posted by inventor2014
i feel like im confusing myself, i thought 3400 mAh is 3.4 Ah right?
Yes.

so that's 3.4 A per hour,
No. You just said up top that it's 3.4Ah, not 3.4A/h. Look very carefully. The lack of a division symbol is important. The unit is ampere-hour, not ampere/hour. Think about what would be implied by the latter. The longer you pull current, the more amps you automagically get? I wish it were so.

The correct unit is ampere-hour, which means, logically enough, that the more current you draw, the shorter the time you can draw it. The product of the current and duration should be roughly constant (although, as others have posted, that isn't quite true, either. You can't pull a billion amperes from an AAA battery ever).

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