Thermal expansion of brass balls
Task number: 2115
A brass sphere has a diameter of 4 cm at 15 °C. How much do you need to increase its temperature in order for the sphere not to fit in a hole of 4.04 cm in diameter?
Hint
To calculate the change in diameter with temperature, you need to know the so-called linear thermal expansion coefficient α, which you can find in the tables.
Analysis
The sphere does not pass through the hole if its diameter increases to more than 4.04 cm. In order to determine the dependence of the diameter of the sphere on the temperature, we need to know the brass coefficient of linear thermal expansion α, which we find in the tables. We express the change of the temperature from the formula for thermal expansion and we calculate it using the appropriate values then.
Notation
t0 = 15 °C Initial temperature of the sphere d0 = 4 cm The initial sphere diameter d1 = 4.04 cm Hole diameter Δt = ? Temperature change From the tables:
α = 20·10−6 °C−1 Linear thermal expansion coefficient Solution
We have to determine at what temperature the diameter of the sphere will be equal to the diameter of the hole. In order to determine the dependence of the diameter on the temperature, we need to know the linear thermal expansion coefficient α (we are considering the change in one dimension, therefore we are working with the linear expansion coefficient).
\[d_1=d_0\left(1+\alpha\mathrm{\Delta}t\right)\]We can express the desired temperature change Δt:
\[\frac{d_1}{d_0}-1=\alpha\mathrm{\Delta}t\] \[\mathrm{\Delta}t=\frac{1}{\alpha}\left(\frac{d_1}{d_0}-1\right)\]Numerical substitution:
\[\mathrm{\Delta}t=\frac{1}{20\cdot{10^{-6}}}\left(\frac{4.04}{4}-1\right)\,{\,^{\circ}\mathrm{C}}=500{\,^{\circ}\mathrm{C}}\]Answer
We must increase the temperature of the brass by at least 500 °C.
Link to an experiment
An experiment for this problem can be found here: Thermal Volumetric Expansion of Metals.
