# Thermal physics problem on steel bar

1. Dec 9, 2007

### veemo

[SOLVED] Thermal physics problem

1. The problem statement, all variables and given/known data

A steel bar 10cm long is welded end-to-end to a copper bar 20cm long. Both bars are insulated perfectly along their sides. Each bar has a radius of 2.0cm. The free end of the steel bar is maintained at 100$$^{o}$$C and the free end of the copper bar is maintained at 0$$^{o}$$C. Find the temperature at the junction between the two bars and the total rate of flow of heat.

k$$_{steel}$$=50.2Wm$$^{-1}$$K$$^{-1}$$
k$$_{copper}$$=385.0Wm$$^{-1}$$K$$^{-1}$$

2. Relevant equations

H = $$\frac{dQ}{dt}$$ = -kA$$\frac{dT}{dx}$$

3. The attempt at a solution

Copper:

$$\frac{dQ}{dt}$$ = $$\frac{-kA(373-T)}{L}$$ = $$\frac{-385\pi(2X10^{-2})^{2}(273-T)}{20X10^{-2}}$$

Where T is the point at the junction between the bars, piXr$$^{2}$$ has been substituted in for A, and 273K = 0C

so $$\frac{dQ}{dt}$$ = -2.42(273-T)

Steel:

$$\frac{dQ}{dt}$$ = $$\frac{-50.2\pi(2X10^{-2})^{2}(373-T)}{10X10^{-2}}$$

As 373K = 100C

so $$\frac{dQ}{dt}$$ = -0.63(373-T)

I then said H$$_{copper}$$ = H$$_{steel}$$ which I'm not sure is correct as in the example I was given in class the two bars were of equal length, but continued with the following:

2.42(273-T)=0.63(373-T)
2.42T-0.63T=660.4-235.3

T=$$\frac{425.1}{1.79}$$= 238K

I'm not sure this is right as it is less than the temperature of the free end of the copper bar and surely the result should be between 273K and 373K?
Also I don't know where to start on finding the total rate of flow of heat... The only thing I could think of was adding the individual heat flows together but I'm sure there's more to it than that.

Last edited: Dec 10, 2007
2. Dec 10, 2007

### rl.bhat

2.42(273-T)=0.63(373-T) This step is wrong. Rewright as 2.42(T- 273)=0.63(373-T) Now solve fro T.

3. Dec 10, 2007

### veemo

Fantastic, thanks! 294K, that sounds better =)
How would I start going about finding the total rate of flow of heat?
I'm still not sure what to use for that.

4. Dec 10, 2007

### rl.bhat

Rate of flow of heat through the compound conductor is given by
delta(Q)/delta(t) = A*delta(T)/( l1/k1 + l2/k2)

5. Dec 10, 2007

### veemo

Brilliant, thanks, I can't find that in my lecture notes -anywhere-.
I'll try it.

6. Dec 10, 2007

### veemo

Ok I got 50.0W, that sounds sensible.
Thanks again!