A cylindrical aluminium can holds 500 cm^3 of water, with the can's...

Initial volume of water, $V_{0} = 500 \text{ cm}^{3}$

Initial area of cross-section of the can, $A_{0} = 125 \text{ cm}^{2}$

Initial height of the water level, $h_{0} = \dfrac{V_{0}}{A_{0}} = \dfrac{500}{125} = 4 \text{ cm}$

Change in temperature, $\Delta T = 80^{{}^{{}^{\circ}}}\text{C} - 10^{{}^{{}^{\circ}}}\text{C} = 70^{{}^{{}^{\circ}}}\text{C}$

After the temperature is raised, the new volume of water is:
$V = V_{0}(1 + \gamma \Delta T)$

The new inner cross-sectional area of the aluminium can is:
$A = A_{0}(1 + 2\alpha \Delta T)$

The new height of the water level is:
$h = \dfrac{V}{A} = \dfrac{V_{0}(1 + \gamma \Delta T)}{A_{0}(1 + 2\alpha \Delta T)} = h_{0}\dfrac{1 + \gamma \Delta T}{1 + 2\alpha \Delta T}$

Since $2\alpha \Delta T \ll 1$, using binomial approximation:
$h \approx h_{0}(1 + \gamma \Delta T)(1 - 2\alpha \Delta T) \approx h_{0}(1 + (\gamma - 2\alpha)\Delta T)$

The rise in the water level is:
$\Delta h = h - h_{0} = h_{0}(\gamma - 2\alpha)\Delta T$

Given:
\gamma = 3.2 \times 10^{-4} ^{{}^{{}^{\circ}}}\textC^{-1} = 320 \times 10^{-6} ^{{}^{{}^{\circ}}}\textC^{-1}
\alpha = 23 \times 10^{-6} ^{{}^{{}^{\circ}}}\textC^{-1}

Substituting the values:
\gamma - 2\alpha = 320 \times 10^{-6} - 2(23 \times 10^{-6}) = 274 \times 10^{-6} ^{{}^{{}^{\circ}}}\textC^{-1}

$\Delta h = 4 \times (274 \times 10^{-6}) \times 70$

$\Delta h = 280 \times 274 \times 10^{-6} = 76720 \times 10^{-6} \text{ cm}$

$\Delta h = 0.07672 \text{ cm} \approx 0.076 \text{ cm}$