Power dissipation due to temperature change

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SUMMARY

The discussion centers on calculating power dissipation in a heater when the wire temperature changes from 800°C to 200°C. The initial power dissipation is 500W at 110V, with a resistance calculated to be 24.2 ohms. Using the formula Rf=R0(1+a(T-T0)), the resistance at 200°C is found to be 18.392 ohms, leading to a recalculated power dissipation of approximately 610W. The accuracy of the answer is questioned, with a suggestion to verify with the lecturer due to potential discrepancies in provided solutions.

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Homework Statement



Heater dissipates 500W when applied potential diference is 110V and wire is 800C. How much power would it dissipate if the wire temperature were held at 200C?

Homework Equations



a=4.0e-4

p=V^2/R
Rf=R0(1+a(T-T0) where R is resistance after temp change
R0 is initial resistance
a is coefficient of resistivity

The Attempt at a Solution



P=V^2/R
R=V^2/P

R=110^2/500
R=24.2

Rf=24.2(1+4.0e-4(200-800)
Rf=18.392

P=110^2/18.392


I'm sure it is just a silly mistake I have done somewhere only just started electronics today after a 2 month university break!

Answer should be 610W

Thanks for any help
 
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The method is correct and the calculation is correct. Unless you are expected to use something other than a linear dependence of resistance on temperature (which would be more realistic for a temperature change as big as 400 degrees), the correct answer should be what you have. How do you know the correct answer is 610W?
 
Cheers for the reply, that is what the answer says but it has been known to be wrong sometimes.

I'll double check with my lecturer, he my have just wrote the wrong answer down.

Thanks again
 

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