# Homework Help: Two rods are connected one silver and the other gold.

Tags:
1. Sep 10, 2015

### smashman101

• Missing homework template due to originally being posted in other forum.
Two metal rods, one silver and the other gold, are attached to each other. The free end of the silver rod is connected to a steam chamber, with a temperature of 100oC, and the free end of gold rod to a ice water bath, with a temp of 00 C. The rods are 5.0 cm long and have a square cross-section, 2.0cm on a side. How much heat flows through the two rods in 60s? The thermal conductivity of silver= 417 W/(m*K), and that of gold is 291 W/(m*K). No heat is exchanged between the rods and the surroundings, except at the ends.

The answer is 8.2 KJ, as given by professor. I cannot get that answer.

He gave us a starting point.
H=ΔQ/Δt = kA((TH-TL)/L)
Where TH=100°C, TL=0°C, Ks=417 W/(m*K), Kg= 291 W/(m*K), A=.0004m2, L=.05m, t=60s

You must find the temperature where the two meet first, T'.

Heat loss=Heat gain; Qsilver=Qgold
ksA((TH-T')/L)=kgA(T'-TL)/L)
ksTH-ksT'=kgT'-kgTL
T'=((ksTH)-(kgTL)/(ks+kg))
So...T'=417*100-291*0/(417+291) T'=58.8°C

Then us 58.8°C for the low and high in the initial equation to get heat.
This is were it gets messy for me.
He gave the hint ΔQ=(ΔQ/Δt)*Δt
Hs=ksA((TH-T')/L)
Hs=417*(.0004)((100-58.8)/.05)
Hs=137.4 J/s

Hg=291*(.0004)((58.8-0)/.05)
Hg=136.8 J/s

We know Q=H*t
Qs=Hs*t=137.4*60=8246J⇒8.2KJ
Qg=Hg*t=136.8*60=8213J

ΔQ=(ΔQ/Δt)*Δt
So (Qs-Qg)/60)*60=3322J←Is not right.

I saw that since we found T', and Hg=Hg, the total heat for the system is just Hg or Hg.

Ive tried many ways, it just seems i cannot get 8.2KJ
If anyone can shed some light that would be much appreciated!!
[/SUB][/SUB]

Last edited: Sep 10, 2015
2. Sep 10, 2015

### Staff: Mentor

In general: you should work with units, that often helps to spot mistakes. This problem is one example.

Several brackets are missing. "a-b/c" is "a - (b/c)", but your calculations need "(a-b)/c".
What is 0.004? It is not the cross-section in square meters. 0.005 is not the length in meters either. Fix both and your Qs and Qg values match the given answer, as they should. The small difference between the two numbers comes from rounding errors I guess. You could keep one or two more digits for the temperature in the middle.

I don't know what you are calculating here, you got the answer in the lines above already. It does not make sense to calculate the difference between the thermal energy conducted in silver and gold.

I would not calculate the temperature in the middle - it is faster if you convert the conductivity to thermal resistance, because you can add the resistances of silver and gold then. The approach needs more thought but the calculations are easier.

3. Sep 10, 2015

### smashman101

Ha, I cant believe i just messed up the conversions, i've spent way too long on this one because of it. I kept going with it, trying to to get the answer down a couple digits. I was thinking there had to be a way to go directly from one side to the other without solving for T', but we have yet to learn this thermal resistance you speak of.

4. Sep 10, 2015

### Staff: Mentor

In case you had some introduction to electrical circuits already: it is the same concept there with two resistances in series.

If you take the conductivity values and the sizes of the objects, you can calculate the conductivity in W/K. "What is the heat flow for a given temperature difference". You get two values (one for silver, one for gold), but you want to find a single one for the total system (as you know the total temperature difference).
Instead of conductivity, consider the inverse value (let's call this R). It has units K/W and tells you how large the temperature difference (between the two ends) is for a given heat flow. Those temperature differences add up - you know their sum is 100.

Therefore, Rs + Rg = Rtotal

Take the inverse value again, and you have the conductivity in W/K for the total system. Multiplied by 100 K you get the total heat flow.