Heating coil

  • Thread starter pat666
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  • #1
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Homework Statement


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.


Homework Equations


ρ=ρ0 [1+α(T-T0 ) ]


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.
 

Answers and Replies

  • #2
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I think that "Resistivity of nichrome at 273 K = 1.08 x 10-6 Ωm." is meant to say 373K
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.
 
  • #3
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Yeah it could be that - the question is not written well at all I will ask tomorrow and find out.
 
  • #4
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turns out the question is right.
 
  • #5
rock.freak667
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Homework Equations


ρ=ρ0 [1+α(T-T0 ) ].
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 ?
 
  • #6
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p=1.08 x 10-6[1+8.0 x 10-5(373-273)]
will that work??
 
  • #7
rock.freak667
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p=1.08 x 10-6[1+8.0 x 10-5(373-273)]
will that work??
Yes that will give you ρ. Now how does ρ relate to the length of wire, resistance and cross-sectional area?
 
  • #8
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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.
 
  • #9
rock.freak667
Homework Helper
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Ive found resistivity at 373K to be 1.089*10^-6.
then R=p(L/A)
Should be correct assuming you calculated correctly (I didn't check the number)

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
This should be correct.
 
  • #10
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ok thankyou, as long as the procedure is right.
 
  • #11
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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.
 
  • #12
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what do you mean??? it all checks out - are you talking sig figures or something?>
 
  • #13
58
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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:

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