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smashman101
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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]
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]
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