Question

Two metal rods of equal length-one aluminum, the other stainless steel-are connected in parallel with a temperature of 10.0 ∘C at one end and 122 ∘C at the other end. Both rods have a circular cross section with a diameter of 7.00 cm Part A Determine the length the rods must have if the combined rate of heat flow through them is to be 17.5 J per second. Express your answer using two significant figures.

Answer 1

By fouriers law

∆Q / ∆t = -k A ∆T / ∆x

∆Q / ∆t = heat / time
k = conductivity of the specific material
A = cross sectional area
∆T / ∆x = change in temperature of the two ends / length of the rod

Now you have two rods in parallel.. and like resistance, the total heat transfer goes like this...

1 / Qt = 1 / Q1 + 1 / Q2
so that 1 / (17.5 J/s) = [∆x / (kAl x A x ∆T) ] + [ ∆x / ( kSS x A x ∆T) ]

and...
∆T = 112.0°C
KAl = 237 J/(s m K)
KSS = 45 J/(s m K).
A = pi x D² / 4 = 3.848x10^-3 m²

1 / (17.5 J/s) = [∆x / (237 J/(s m K) x (3.848x10^-3 m²) x 112.0°C) ] + [ ∆x / ( 45 J/(s m K) x (3.848x10^-3 m²) x 112.0°C) ]

1 / (17.5 J/s) = [∆x / (102.14 J m / s)] + [ ∆x / ( 19.394 J m / s) ]

(1980.9 J m / s) / (17.5 J m / s) = 19.394∆x + 102.14∆x

x = 0.9314 m = 93.14 cm