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 1000C, and the free end of gold rod to an ice water bath, with a temperature of 00C. The rods are 5.0 cm long and have a square cross-section, 2.0 cm on a side. How much heat flows through the two rods in 60 s? The thermal conductivity of silver is 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.
A) 8.2 KJ
B) 9.5 KJ
C) 12 KJ
D) 14 KJ
E) 16 KJ
Correct Answer is option A, i.e. 8.2KJ
Solution:- 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. 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
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