Answer to Question #86279 in Molecular Physics | Thermodynamics for Jason

Question

The diameter of the round brass rod is 30.05 mm and the steel ring the inner diameter is 30.00 mm at 20 ° C. At what temperature the steel ring must be heated to fit the brass rod inside?

Answer 1

Solution:

Change in length of the ring with temperature increasing:

$$\Delta{L}=L_0\alpha\Delta{T}$$

Then change in temperature:

$$\Delta{T}=\frac{\Delta{L}} {L_0\alpha}=\frac{L-L_0} {L_0\alpha}$$

Length of the ring after temperature increasing:

$$L=2\pi R=\frac{2\pi D} 2=\pi D$$

Analogously we can find the length of the ring before temperature increasing:

$$L_0=\pi D_0$$

Therefore:

$$\Delta{T}=\frac{\pi D-\pi D_0} {\pi D_0\alpha}=\frac{D-D_0} {D_0\alpha}$$

Coefficient of linear expansion for steel:

$$\alpha=1.2\times 10^{-5} \: (1/^\circ C)$$

Let's calculate the change in temperature:

$$\Delta{T}=\frac{30.05-30.00} {30.00\times 1.2\times 10^{-5}}=139\:(^\circ\:C)$$

Therefore  the temperature the steel ring must be heated to fit the brass rod inside:

$$T_2=T_1+\Delta{T}=20^\circ\:C+139^\circ\:C=159\:^\circ\:C$$

Answer: $T_2=159\:^\circ\:C$.