Help! Solving a Physics Problem - Can You Assist?

[suf7]

Physics Problem!

What do i do with this question?...can sum1 help?

The element of an electric fire consists of 5m of wire of 0.5mm diameter. The resistivity of the wire at 150C is 1.12 * 10^-6 Ωm. When connected to a 240V supply the fire dissipates 2 kW and the temperature of the element is 1015 C. Determine a value for the mean temperature co-efficient of resistance of the wire??

 

[Galileo]

The energy dissipation in a wire is:
$$P=VI=\frac{V^2}{R}$$,
this allows you to get the resistance of the wire at 1015 deg. celsius.
You can also calculate the resistance of the wire at 150 C, given its length, diameter and resistivity.

Now use the definition of the temperature coeff. of resistance.

 

[suf7]

sorry, i don't really understand...what do each of the letters in the formula represent?...im new to physics and don't really know what to do.

 

[suf7]

im still stuck with this?

 

[Galileo]

Read your textbook about this.
They may use different letters, but most of them are applied universally.
'P' is power, measured in Watts
'V' is the potential difference, measured in Volts
'I' is the current, in Amperes.
'R' is resistance, in Ohms.

What I was saying was; the energy dissipated in a wire every second is equal to VI, or V/R^2.

I hope that's what you were asking...

 

[suf7]

Read your textbook about this.
They may use different letters, but most of them are applied universally.
'P' is power, measured in Watts
'V' is the potential difference, measured in Volts
'I' is the current, in Amperes.
'R' is resistance, in Ohms.

What I was saying was; the energy dissipated in a wire every second is equal to VI, or V/R^2.

I hope that's what you were asking...

i have tried reading the textbook but it is VERY confusing, it doesn't make any sense??...the formula you have stated, i don't understand where does temperature go into it??

 

[Galileo]

i have tried reading the textbook but it is VERY confusing, it doesn't make any sense??...the formula you have stated, i don't understand where does temperature go into it??

Nowhere explicitly. (This formula is not in your book?)
But it's the resistance that is a function of temperature.

The idea is: Calculate the resistance of the wire at 150 C, and calculate the resistance of the wire at 1015 C. Then use this to find the average (or mean) temperature coefficient.

Calculating the resistance at 150 C:
Use the formula:
$$R=\frac{\rho L}{A}$$
Where $\rho$ is the resistivity, L the length and A the cross sectional area of the wire.

Calculating the resistance at 1015 C:
Use the equation from before (but solve for R):
$$R=\frac{V^2}{P}$$

 

[suf7]

Nowhere explicitly. (This formula is not in your book?)
But it's the resistance that is a function of temperature.

The idea is: Calculate the resistance of the wire at 150 C, and calculate the resistance of the wire at 1015 C. Then use this to find the average (or mean) temperature coefficient.

Calculating the resistance at 150 C:
Use the formula:
$$R=\frac{\rho L}{A}$$
Where $\rho$ is the resistivity, L the length and A the cross sectional area of the wire.

Calculating the resistance at 1015 C:
Use the equation from before (but solve for R):
$$R=\frac{V^2}{P}$$

Ok, but what is the capital P and V^2 in the bottom formula?...im confusing myself?

 

[Galileo]

..maybe you're thinking too hard...
$$R=\frac{V^2}{P}$$
and
$$P=\frac{V^2}{R}$$
are one and the same formula. I just wrote it this way since P and V are given in the problem.

 

[suf7]

..maybe you're thinking too hard...
$$R=\frac{V^2}{P}$$
and
$$P=\frac{V^2}{R}$$
are one and the same formula. I just wrote it this way since P and V are given in the problem.

oh rite...lol... thanks for the help!