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Design details for a butter knife warmer

  1. Sep 24, 2004 #1
    Hi everyone!
    I need help on how to build a butter knife that gets warm. I know I need to put a nichrome wire in the blade but would anything go at the end? Would I connect the wire to a battery? How do I find out what size battery to use and make it so the blade gets to a specific heat?

    Any help would be greatly appreciated.

    ~Thanks
     
    Last edited: Sep 24, 2004
  2. jcsd
  3. Sep 25, 2004 #2
    batteries wont last long
    nichrome wire might not work at low voltage,there
    are other resistance/heating wires available for low voltage applications
     
  4. Sep 25, 2004 #3
    i suggest you consult an electrician about this cause it could get
    dangerous if you dont know what youre doing!!!!!!!!
     
  5. Sep 26, 2004 #4

    Ivan Seeking

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    You need to read this.

    http://www.sci-journal.org/index.ph.../v3n1k44.html&link=reports/home.php&c_check=1

    and this
    http://www.the12volt.com/ohm/ohmslaw.asp

    By controlling the watts [power produced] you will control the heat produced. More watts means more heat. You don't want to mess with AC power [110 volts] especially without help from an adult who is knowledgeable about electricity.

    I was thinking that you may not even need the nichrome wire. Maybe you could just use a resistor. Why don't you try this as a place to start. Get a 10 ohm, 1/4 watt resistor. You can get these at Radio Shack. Use one D cell to power the resistor. This will overload the battery a bit so don't run it very long; say no longer than one, five, ten, then thirty seconds at a time at first. Check the battery frequently and make sure that it doesn't get hot. You will probably see the battery voltage drop as a result of the overload. The resistor will get hot and should produce about 0.2 watts of power. See if this will do the job. Then you can think about your next step.

    You should work with an adult who is knowledgeable about electricity. For one, the battery can overheat and even explode. Always wear safety goggles.

    EDIT: Does anyone know the internal resistance [ESR] of a typical D cell?
     
    Last edited: Sep 27, 2004
  6. Sep 29, 2004 #5
    Would I have the end of the resistor just touching the battery and the other end not touching anything?

    The D cell has an internal resistance of about 3 Ω
     
    Last edited: Sep 29, 2004
  7. Sep 29, 2004 #6

    Cliff_J

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    Shay the resistor is the heating element. It needs to have current passing through it to heat up. So the resistor is on the butter knife and then wires connect each end of it to the battery.

    To find the power, consult ohm's law. Use algerbra to solve the following equasions:

    Voltage = Current * Resistance
    Power = Voltage * Current

    Then you could weigh the butter knife and find the grams of steel in it. Now what is the heat capacity of the steel in the butterknife? How about the butter that will coat the blade (could probably use water to get close) as its cutting? That should be in your report as to how you determined the resistance and power level.

    Where did you get the ESR figure from for a D cell? Sounds wayyyyyyy too high as I found 3 miliohms in a quick google search.
    Cliff
     
  8. Sep 29, 2004 #7
    Since I'm using a D cell battery how would I connect the wires to the battery? For example if the D cell battery was in the battery compartment of a flash light the ends of the battery would be touching the metal spring and the metal plate. Would I connect the resistor wires to the spring and the plate?

    Would it work with a D cell battery? Is there any way I could use a AA battery?

    The resistor will be in the middle of two blades put together to make one blade. Will the resistor have to touch anything while in the blade? Since both ends of the resistor have to touch the battery the resistor will touch the battery, go up the blade, turn around, and touch the other end of the battery?

    I found the ESR from a google search too. Mine could be wrong.
     
    Last edited: Sep 29, 2004
  9. Sep 29, 2004 #8

    Cliff_J

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    Battery sizes roughly give a clue of the power available.

    You'll notice an AAA, AA, C, D battery are all 1.5V so voltage doesn't change. Instead its the current they can deliver that changes (determined by the ESR) and how long they can deliver that current (determined by the capacity).

    Look at this chart:
    http://www.techlib.com/reference/batteries.html

    According to this if you use a D cell and used 12 amps (12,000 miliamps) it should last 1 hour, or if you used 6 amps it should last 2 hours. Mathimatically that's correct but remember our ESR from above? It will drastically reduce that and the more current you pull the less accurate the formula. A lot of batteries are rated for capacity at 20 hours so capacity at 1 hour its probably overrated by 5-10x. So with 12A from a D cell I'd guess maybe 6-12 mins before its dead and I might be optimistic.

    Ok, back up from the construction for a second. Answer these questions:
    How much heat power do you need?
    How much voltage do you have with 1 cell? How about 2 cells?
    How much resistance do you need to flow the correct amount of current from that voltage?
    How long will each battery last flowing that current?

    Yes, you will need the resistor to touch the blade. You'll likely want to cover the resistor leads with heatshrink or electrical tape. Then maybe some heatsink compound to assist in the heat transfer from resistor to blade.

    Oh, and make sure the resistor has a rating higher than the number of watts you plan to use.

    Cliff
     
  10. Sep 29, 2004 #9

    Ivan Seeking

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    Thanks for jumping in Cliff. I am fighting time right now...darned work! I have PF to attend to. :biggrin:

    But not so much higher that the resistor never gets hot [hence the 0.2 watt target]. Also, the knife will heat sink the resisitor so I was thinking that with good thermal coupling he may be able to overdrive things a bit. I was thinking that staying just below the rated power is a good place to start in order to insure a high max temperature.

    Note: I just wanted to be sure that we weren't close to the ESR with a 10 ohm resistor. I thought it was more like 3 ohms for a D cell...hmmm...learn something new every day. Now, wait, that can't be right. At 1.5 volts and an ESR of milliohms, the total current available is way too high.
     
    Last edited: Sep 29, 2004
  11. Sep 30, 2004 #10
    Is this correct?:
    ~~~~~~~

    Since the specific heat of silver, cal/g * degrees C, is .056 you change that into watts which is .234 W. So is .234 W the amount of heat power I need?
    ~~~~~~~

    With one cell I have 1.5 volts, and with two clees I have 3 volts.
    ~~~~~~~

    P=VC
    .234= 1.5C
    C=.156

    V=CR
    1.5=.156R
    R=9.615

    For 1.5 V I need 9.615 ohms of resistance?
    ~~~~~~~

    P=VC
    .234=3C
    C=.078

    V=CR
    3=.078R
    R=38.462

    For 3 V I need 38.462 ohms of resistance?
    ~~~~~~~

    With 1.5 V the battery will last 77 hours?
    With 3 V the battery will last 154 hours?
    ~~~~~~~
    Is any of this correct?
     
  12. Sep 30, 2004 #11

    Cliff_J

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    Good, you looked up specific heat. That was the point of the question, you can describe that different materials heat differently. You'll likely want to use steel instead of silver. I did a quick search and this site came up, pretty handy comparison:
    http://www.ex.ac.uk/trol/dictunit/notes5.htm

    Note that the level for water is much higher. You need to factor in how long you can wait and how much heat will be transmitted to the air from the knife to be exact, unnecessary in all likelyhood. But if instead you could heat up the butter knife first and then cut the butter you can take advantage of the thermal mass of the butter knife. Lets assume .2W is a nice starting point, its a safe level to begin experimenting.

    Your math looks fine for all the rest. (As a side note, on the internet most people will use the standard letters E=voltage and I=current and R=resistance. Its a long story why....)

    You can find resistors of sizes close to those values, like the 10 ohm mentioned by Ivan above or a 39 ohm resistor if you choose to use two cells. As you can also tell now that you've done the math (and should have a spiffy report with) you can see a 1/4 watt resistor will work fine.

    As another side note, you can see that for the same power level a doubling of voltage means 4x the resistance is need to stay at the same power level. Or if you double the voltage and keep the resistance the same, you have 4x the power.

    -----------------------------------------

    Ivan - 3 ohms still sounds too high, even at .2A the drop is .6V? Sure .003 sounds way too low (how good is a unknown source on the internet) but maybe somewhere in the middle like 1 ohm? Maybe find time to crack open a flashlight this weekend and test a couple batteries. :smile:

    Cliff
     
  13. Sep 30, 2004 #12

    Ivan Seeking

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    A couple of more details here. The specific heat is in calories per gram per degree centigrade. So, after converting calories to joules [not watts], you need 0.234 joules, per gram of silver, per degree C of temperature change for the silver mass.

    Now, power - watts - is the number of joules of energy per second. The 0.234 watts calculated would raise one gram of silver, one degree C per second. Two grams of silver would heat at a rate of 0.5 degrees C per second, and so on. Also, as Cliff pointed out, this does not account for the heat lost.

    Ideally we would want to include the mass and specific heat of the butter as well. This, plus the heat lost to the atmosphere would give us the total heat load, but you can probably ignore this for now. It should be mentioned in your report though.
     
    Last edited: Sep 30, 2004
  14. Sep 30, 2004 #13

    Ivan Seeking

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    If we see you selling this on QVC, Cliff and I will be waiting for our royalty checks. :biggrin:

    Oh yes, there is another issue of how fast the resistor tranfers heat to the metal. As mentioned earlier, use heat sink paste [available at Radio Shack], but also make sure that the resistor [not the leads] is in direct contact with the metal blade. This will be a little tedious to set up properly but it is very important.


    You may also find that the metal blade is able to dissipate the heat [hence power] faster than you can supply it. If this happens you will need to produce more power.
     
    Last edited: Sep 30, 2004
  15. Sep 30, 2004 #14
    I get around this 'hard butter' problem by avoiding refrigeration.

    Good luck
     
  16. Sep 30, 2004 #15

    Ivan Seeking

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    Actually, if you use real butter you can get minor food poisoning this way. Most such cases are misinterpreted as minor influenza or just a bad stomach.
     
  17. Sep 30, 2004 #16
    You mean it wasn't really the flu making my belly hurt all these years?

    I hadn't known that, but generally use the imitation stuff, if any at all.
     
  18. Sep 30, 2004 #17

    megashawn

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    hey, instead of making a heated knife, why not make a knife heater?

    You know, some lil thing just barely wide enough to slide a standard butter knife in it. Then, use some heating elements to warm the knife. This way, you can warm any knife, instead of one special knife.
     
  19. Sep 30, 2004 #18

    Ivan Seeking

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    I hate to guess what I've spent on Kaopectate due to my long, room temp butter habbit. It took me years to make the connection. I only realized the potential connection when I saw a woman from the National Centers for Disease Control who stated that something like 80% of all minor, alleged cases of influenza are really minor food poisoning.
     
  20. Oct 1, 2004 #19
    If I use two D cell batteries an tthe 39 ohm resistor will the blade heat up faster?

    The 10 ohm .25 watt resistor is short in length. How would I have both ends of it touching the battery if it is soshort. Would I connect some kind of wire t theend of the resistor to make it longer?
     
  21. Oct 1, 2004 #20

    Ivan Seeking

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    No. In this case the power is about the same. Just try multiplying the voltage and current for each. In order to increase the power you would need less resistance or more voltage. Note that you can also get 1/2 watt resistors and higher.

    Yes. Make sure that you don't cause a short circuit. You might want to consider using an old flashlight for a battery holder.
     
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