The volume of a 1 cm\(^3\) metal ball increases by 0.0018 cm\(^3\) when heated through temperature θ. If the linear expansivity of the ball is 2.0 x 10\(^{-...

Volume expansion is governed by the cubic expansivity \(\gamma\): \[\Delta V = V_1 \gamma\, \Delta\theta.\] For a solid the three expansivities are related by \(\gamma = 3\alpha\) and \(\beta = 2\alpha\), because a solid expands by the same fractional amount in each of its three perpendicular directions. Here the linear expansivity is given, so convert first: \[\gamma = 3\alpha = 3 \times 2.0\times10^{-5} = 6.0\times10^{-5}\,\text{K}^{-1}.\]

Now substitute the data, with \(V_1 = 1\,\text{cm}^3\) and \(\Delta V = 0.0018\,\text{cm}^3\): \[0.0018 = 1 \times 6.0\times10^{-5} \times \theta \quad\Rightarrow\quad \theta = \frac{0.0018}{6.0\times10^{-5}} = 30.\] The temperature rise is \(30\) kelvin, which is a rise of \(30\,^\circ\text{C}\). A change of temperature has the same numerical value on both scales because the degree sizes are identical, which is why an expansivity quoted in \(\text{K}^{-1}\) may be used directly with a Celsius temperature change.

The mistake that produces \(90\) is using \(\alpha\) itself in the volume formula, and the mistake that produces \(45\) is using \(\beta = 2\alpha\), the area expansivity. Match the expansivity to the dimension being measured: length with \(\alpha\), area with \(2\alpha\), volume with \(3\alpha\). Note also that only the ratio \(\Delta V / V_1\) matters, so the units of volume cancel and no conversion of cubic centimetres is needed.