A wine barrel has a diameter at its widest point of \(134.460 \mathrm{cm}\) at a temperature of \(20.0^{\circ} \mathrm{C} .\) A circular iron band, of diameter \(134.448 \mathrm{cm},\) is to be placed around the barrel at the widest spot. The iron band is \(5.00 \mathrm{cm}\) wide and \(0.500 \mathrm{cm}\) thick. (a) To what temperature must the band be heated to be able to fit it over the barrel? (b) Once the band is in place and cools to \(20.0^{\circ} \mathrm{C},\) what will be the tension in the band?

First, we need to determine the difference between the diameter of the barrel and the diameter of the circular band. To do this, subtract the diameter of the band from the diameter of the barrel: Δd = d_barrel - d_band Δd = 134.460 cm - 134.448 cm Δd = 0.012 cm So, the band needs to expand by 0.012 cm in diameter to fit over the barrel.

Next, using the coefficient of linear expansion (\(α\)) for iron, we can find the temperature change required to expand the band by 0.012 cm. The linear expansion formula is: \(ΔL = L₀ * α * ΔT\) Where ΔL is the change in length (or diameter in this case), L₀ is the initial diameter, α is the coefficient of linear expansion, and ΔT is the change in temperature. We rewrite the formula to find ΔT: \(ΔT = \frac{ΔL}{L_0 * α}\) The coefficient of linear expansion for iron is roughly \(α = 12 × 10^{-6} K^{-1}\). \(ΔT = \frac{0.012 cm}{ 134.448 cm * 12 × 10^{-6} K^{-1}}\) \(ΔT ≈ 74.2 K\) So, the iron band needs to be heated to 94.2°C to fit over the barrel (20°C + 74.2°C).

Now that the iron band is heated and fits the barrel, we need to determine the change in its circumference. The circumference formula is: \(C = πd\) The change in circumference can be found by taking the difference between the heated circumference and the initial circumference: \(ΔC = C_{heated} - C_{initial}\) \(ΔC = π(d_{initial} + Δd) - πd_{initial}\) \(ΔC = π(134.460 cm) - π(134.448 cm)\) \(ΔC ≈ 0.0377 cm\)

To calculate the tension in the band, we use Hooke's law. The formula relating stress, strain, and Young's modulus for a material is: \(Stress = Strain × Y\) Where Y is Young's modulus for iron (\(Y ≈ 20 \times 10^{10} N/m^2\)). Stress, in terms of force and cross-sectional area, is: \(Stress = \frac{Force}{Area}\) And strain, in terms of the change in length and the initial length, is: \(Strain = \frac{ΔL}{L₀}\) Now, we need to find the cross-sectional area of the band: \(Area = width * thickness\) \(Area = 5.00 cm * 0.500 cm\) \(Area = 2.5 cm^2\) We convert the area to square meters (1 cm² = 0.0001 m²): \(Area = 2.5 * 0.0001 m^2\) \(Area = 0.00025 m^2\) Now, we can find the force (tension) in the band. Since the circumference of the band has changed, we can relate stress and strain to force, area, and circumference: \(Force = \frac{ΔC}{C_0} * Area * Y\) \(Force = \frac{0.0377 cm}{134.448 cm} * 0.00025 m^2 * 20 \times 10^{10} N/m^2\) \(Force ≈ 1400 N\) So, the tension in the band when cooled to 20°C is approximately 1400 N.