A steel sphere sits on top of an aluminum ring.The steel sphere (α = 1.10 × 10−⁵/C°)has a diameter of 8 000 cm at 0°C.The aluminum ring (α = 2.40 × 10−⁵/C°)has an inside diameter of 7 991 cm at 0°C.Closest to which temperature given will the sphere just fall through the ring?

Question

Quizplus · Accepted Answer

For the steel sphere to fall through the aluminum ring, the diameter of the sphere must become equal to or slightly less than the inside diameter of the ring as they are heated. The change in diameter for each can be calculated using the formula for linear expansion, ΔL = αL₀ΔT, where ΔL is the change in length (or diameter in this case), α is the coefficient of linear expansion, L₀ is the original length (or diameter), and ΔT is the change in temperature.For the steel sphere:ΔD_steel = α_steel * D_steel_0 * ΔT= 1.10 × 10^(-5) /°C * 8000 cm * ΔTFor the aluminum ring:ΔD_aluminum = α_aluminum * D_aluminum_0 * ΔT= 2.40 × 10^(-5) /°C * 7991 cm * ΔTFor the sphere to just fall through the ring, the final diameter of the sphere should be equal to or less than the final diameter of the ring. Thus, we set the initial diameter plus the change in diameter of the sphere equal to the initial diameter plus the change in diameter of the ring and solve for ΔT.8000 cm + (1.10 × 10^(-5) /°C * 8000 cm * ΔT) = 7991 cm + (2.40 × 10^(-5) /°C * 7991 cm * ΔT)Solving this equation directly for ΔT might be complex without a calculator, but understanding the concept that the aluminum ring expands at a faster rate than the steel sphere due to its higher coefficient of linear expansion, we can infer that the temperature must be high enough to allow for this differential expansion to occur. Given the options, we can estimate or calculate the exact temperature, which in this case, aligns closely with option C, 87°C, based on the coefficients of expansion and the initial dimensions provided.

A Steel Sphere Sits on Top of an Aluminum Ring