Two metal rods, one silver and the other gold, are attached to each other. The f

Question

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

Explanation / Answer

Correct Answer is option A, i.e. 8.2KJ

Solution:- H=Q/t = 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. Q=(Q/t)*t

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=8246J8.2KJ

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