A square brass plate, \(8.00 \mathrm{cm}\) on a side, has a hole cut into its center of area \(4.90874 \mathrm{cm}^{2}\) (at $\left.20.0^{\circ} \mathrm{C}\right) .$ The hole in the plate is to slide over a cylindrical steel shaft of cross-sectional area \(4.91000 \mathrm{cm}^{2}\) (also at \(20.0^{\circ} \mathrm{C}\) ). To what temperature must the brass plate be heated so that it can just slide over the steel cylinder (which remains at \(20.0^{\circ} \mathrm{C}\) )? [Hint: The steel cylinder is not heated so it does not expand; only the brass plate is heated. \(]\)

We are given the following information: - Brass plate dimensions: side length is 8.00 cm, hole's area is 4.90874 cm² - Steel shaft cross-sectional area: 4.91000 cm² - Brass and steel initial temperatures: 20.0°C - Coefficients of linear expansion for brass (\(\alpha_B\)): \(19 \times 10^{-6}/\mathrm{C}^\circ\)

Since the hole is square, we can find its initial side length using the area formula: Side length of hole = \(\sqrt{Area}\) Side length of hole = \(\sqrt{4.90874 \mathrm{cm}^2}\) Side length of hole = 2.215 cm

The change in area due to the change in temperature is given by the formula: \(\Delta A = 2 \alpha A_0 \Delta T\) The final hole area, \(A_h\), can be expressed as the initial hole area \(4.90874 \mathrm{cm}^2\) plus the change in the area \(\Delta A\): \(A_h = 4.90874 \mathrm{cm}^2 + 2\alpha_B (4.90874 \mathrm{cm}^2)(T-20)\)

To slide over the steel shaft, the hole area must be equal to or greater than the cross-sectional area of the steel shaft, which is 4.91000 cm². So, \(A_h \geq 4.91000 \mathrm{cm}^2\)

We can now set the expression for the final hole area equal to 4.91000 cm² and solve for the temperature: \(4.91000 \mathrm{cm}^2 = 4.90874 \mathrm{cm}^2 + 2\alpha_B (4.90874 \mathrm{cm}^2)(T-20)\) Rearrange the equation and solve for temperature \(T\): \(T = 20 + \frac{4.91000 \mathrm{cm}^2 - 4.90874 \mathrm{cm}^2}{2\alpha_B (4.90874 \mathrm{cm}^2)}\) Now plug in the given value of \(\alpha_B = 19 \times 10^{-6}/\mathrm{C}^\circ\) and solve for \(T\): \(T = 20 + \frac{4.91000 \mathrm{cm}^2 - 4.90874 \mathrm{cm}^2}{38 \times 10^{-6} \mathrm{C}^{\circ}^{-1} \cdot 4.90874 \mathrm{cm}^2}\) \(T \approx 30.55^{\circ}C\) Thus, the brass plate must be heated to approximately \(30.55^{\circ}C\) to slide over the steel shaft.