Lead and Silver with different temperatures. Can you guys help me with this problem? Thanks!?

Question

A bar of lead (bar A) is in thermal contact with a bar of silver (bar B) of the same length and area (Fig. P11.59). One end of the compound bar is maintained at Th = 86.0°C while the opposite end is at 30.0°C. Find the temperature at the junction when the energy flow reaches a steady state.
________°C
http://www.webassign.net/sercp/p11-59alt.gif

Answer 1

We know that q = k A dT / s
where
q = heat transferred per unit time (W )
A = heat transfer area (m^2)
k = thermal conductivity of the material (W/m.K)
dT = Temperature difference across the material ( K) = T(hot) - T(cold)
s = material thickness (m)

Since heat transferred per unit time at the junction is the same

k(Pb)A (T2-T1) / s = q = k(Ag) A (T1-T3) / s

k(Pb)=(T2 - T1) = k(Ag)(T1-T3)
finally
T1= [k(Pb)T2 + k(Ag)T3]/[k(Pb)+ k(Ag)]
T1=[35 (86.0 + 273.15 ) + 429 (30.0 + 273.15)]/[35+429]=
T1= 307.4 K or
T1= 34.2 C

Answer 2

Best Answer - Chosen by Asker

..........L...E....A...D> >>Q O========OO-------------O >>Q
.........A.................. J ...................B
.. 86C^........... .T C^............ 30 C^
........<--- L -------> J <---- L ------>
J =metal junction

rods are connected in series, when in steady state, rate of heat flow (like current in series) will be same

Q = conducted heat /sec = - k A ∆T/∆x
------------------------------...
k = thermal conductivity, A=cross-section area=A
∆T = fall in temp = - (T-hot - T-cold)
∆x = L (+ve x)
------------------------------...
(Q)lead = (Q)sil
k(l) A (∆T)A-J / L = k(s) A (∆T)J-B / L
------------------------------...
(∆T)A-J = (86-T) C and (∆T)J-B = (T-30)
k(s) = 429 W/m-C and k(l) = 35.3 (do)
------------------------------...
35.3 (86 - T) = 429 (T-30)
T (429+35.3) = 3035.8+12870 = 15905.80
T = 34.26 deg C
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small change made