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Homework Help: Heating coil

  1. Jul 19, 2010 #1
    1. The problem statement, all variables and given/known data
    A heating coil is to be made, from nichrome wire, which will operate on a 12 V supply and will have a power of 36 W when immersed in water at 373 K. The wire available has a cross-sectional area of 0.10 mm². What length of wire will be required?
    (Resistivity of nichrome at 273 K = 1.08 x 10-6 Ωm.
    Temperature co-efficient of resistivity of nichrome = 8.0 x 10-5 K-1.)
    I think that "Resistivity of nichrome at 273 K = 1.08 x 10-6 Ωm." is meant to say 373K, I will ask tomorrow.


    2. Relevant equations
    ρ=ρ0 [1+α(T-T0 ) ]


    3. The attempt at a solution
    Im thinking that ρ(resistivity) can be changed with P/I^2 then I^2 with V^2/R^2 so i would end up with P*R^2/v^2=ρ0 [1+α(T-T0 ) ] or something like that, I really need some pointers on where to go.
     
  2. jcsd
  3. Jul 20, 2010 #2
    Are you sure the question is not asking you for the length of wire needed in a heating coil, if it is to heat water from 273K (just freezing) to 373K (boiling)? You need a start and an end temp. If you just want to heat water with nichrome then any amount will do...

    Think of some more relevant equations. You need a length in there somewhere to solve for it.
     
  4. Jul 20, 2010 #3
    Yeah it could be that - the question is not written well at all I will ask tomorrow and find out.
     
  5. Jul 20, 2010 #4
    turns out the question is right.
     
  6. Jul 20, 2010 #5

    rock.freak667

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    Homework Helper

    This equation will give you the resistivity at a temperature T. So using the information given and the terms in the equation, what is the resistivity at T=373 K ?
     
  7. Jul 20, 2010 #6
    p=1.08 x 10-6[1+8.0 x 10-5(373-273)]
    will that work??
     
  8. Jul 20, 2010 #7

    rock.freak667

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    Yes that will give you ρ. Now how does ρ relate to the length of wire, resistance and cross-sectional area?
     
  9. Jul 21, 2010 #8
    hey rock.freak667, sorry for the late reply I was at uni for 10hrs today. Ive found resistivity at 373K to be 1.089*10^-6.
    then R=p(L/A)
    V=IR P=IV so P=V/R*V P=V^2/R R=v^2/P im not sure if this is right im just subbing formulas to try and find something useful
    I get
    V^2/P=p(L/A)
    144/36=1.089*10^-6.(L/1.1*10^-7)
    L=0.404m

    Please reply, I have no idea if this is even close to right
    ps-thanks for your help thus far.
     
  10. Jul 21, 2010 #9

    rock.freak667

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    Should be correct assuming you calculated correctly (I didn't check the number)

    This should be correct.
     
  11. Jul 21, 2010 #10
    ok thankyou, as long as the procedure is right.
     
  12. Jul 22, 2010 #11
    hey man the resistivity of nichrome at 273K is given in the question to be 1.08*10^-6ohmm, using the value 1.089*10^-6ohmm gives you a length required that is 4cm greater. dunno if i would trust the lecturer to mark it correctly, even though its correct with your value.
     
  13. Jul 22, 2010 #12
    what do you mean??? it all checks out - are you talking sig figures or something?>
     
  14. Jul 22, 2010 #13
    whoops i read it wrong! i'm still getting 0.36m so i'll update this post when i find the difference

    edit: okay silly error my bad.
     
    Last edited: Jul 22, 2010
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