Question

At 20∘C, the hole in an aluminum ring is 2.200 cm in diameter. You need to slip this ring over a steel shaft that has a room-temperature diameter of 2.204 cm .

To what common temperature should the ring and the shaft be heated so that the ring will just fit onto the shaft? Coefficients of linear thermal expansion of steel and aluminum are

12×10^-6K^-1 and 23×10^-6K^-1 respectively.

Express your answer in degrees Celsius to two significant figures.

Answer 1

The linear thermal expansion is given by the following equation,

ΔL = αLo ΔT

where ΔL is the change in length L, ΔT is the change in temperature, and α is the coefficient of linear expansion,

the linear expansion coefficient of steel is, α = 12 x 10^-6 K^-1 and that of the aluminum is α = 23 x 10^-6 K^-1.

L = Lo(1 + αΔT)

The initial temperature is 20°C. If the final common temperature is T°C, then ΔT = T-20°

At temperature T, the diameter of the hole in the aluminium is:
2.200(1 + 23×10^−6(T-20))

At temperature T, the diameter of the steel shaft is:
2.204(1 + 12×10^−6(T-20))

These must be equal so:
2.200[1 + 23×10^−6(T-20)] = 2.204[1 + 12×10^−6(T-20)]

The common temperature at which the ring and the shaft be heated so that the ring will just fit onto the shaft is,

T = 35650°C = 3600°C = 3.6 ×10^3° C _Ans.(2 significant figures)