Posted by pins on Tuesday, January 29, 2013 at 9:40am.
A cylindrical rod of crosssectional area 5.0mm2 is made by joining a 0.30m rod of silver to a 0.12m rod of nickel. The silver end is maintained at 290K and the nickel end at 440K. given the thermal conductivities of silver and nickel are 0.42kWm1K1 and 91Wm1K1 respectively, calculate:
the temperature of the join under steady conditions, and
b the rate of conduction of heat down the rod
State any assumptions you need to make.

college physics  Elena, Tuesday, January 29, 2013 at 12:20pm
A=5•10⁻⁶ mm²,
Silver: k₁= 420 W/m•K, L₁= 0.3 m,
Nickel: k₂=91 W/m•K, L₂=0.12 m,
T=? q=?
Q/t=q=k•A•ΔT/L
Resistance of the silver rod:
R₁=L₁/k₁A₁ =0.3/420•5•10⁻⁶=142.9 Ω.
Resistance of the nickel rod:
R₂=L₂/k₂A₂ = 0.12/91•5•10⁻⁶= 263.7 Ω.
Two rods are the conductotrs in series:
R=R1+R2=406.6 Ω.
q= ΔT/R= (440290)/406.6=0.369 W.
ΔT₁=q•R₁=0.369•142.9=52.7 K
ΔT₂=q•R₂=0.369•263.7=97.3 K
T=ΔT₁+T₁=52.7+290=342.7 K
or
T= T₂ΔT₂=44097.3=342.7 K
= >temperature of the join is
T=342.7 K ~343 K,
the rate of conduction of heat is
q= 0.369 W ~ 0.37 W
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