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A cylindrical rod of cross-sectional 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 0.42kWm-1K-1 and 91Wm-1K-1 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 -

    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= (440-290)/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= T₂-ΔT₂=440-97.3=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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