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5. A cylinder of diameter 1.0 cm at 30°ct into a hole in a steel plate. The ois to be sto be tlamethole has a diof 0.99970 cm at 30°C. To what tethe plate be heated ? For steel, α = 1.1×10-5 must10 C1

Answer»

Let α be the linear expansion coefficient for a material. Then:

(E1) ΔL / L0 = α * ΔT

Since we are given a linear dimension (the diameter), we can use it, together with (P1) and (E1), to calculate the temperature T. First,

(E2) ΔT = T - T0 = T - 30and(E3) ΔL = L - L0 = 1.00000 - 0.99970 = 0.00030cm

Substituting into (E1):

(E4) ΔL / L0 = α * (T - 30)

= α * T - α * 30

Solving for T:

(E5) T = (ΔL / L0 + α * 30) / α

= (ΔL / L0 ) / α + 30

and substituting from (E3):

= (0.00030 / 0.99970 ) / α + 30

= 0.00030 / α + 30

Now all we need to know is what material the plate is made of, plus an experimentally determined value for α at 30C for that material.

If the material is steel and we assume that

(E6) α30 ≈ α20,

then

(E7) T = 0.00030 / α + 30

= 0.00030 / 13e-6 + 30

= 53.C


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