Question

A steel sphere with radius 1.0010 cm at 41.0°C must slip through a brass ring that has an internal radius of 1.0000 cm at the same temperature. To what temperature must the brass ring be heated so that the sphere, still at 41.0°C, can just slip through? Coefficient of linear expansion α for brass is 19.0 × 10−6 K−1.

Answer 1

Answer:

$\Delta T = 52.6 ^o C$

Explanation:

As we know that radius of the brass ring is given as

$R_{brass} = 1.0000 cm$

radius of the sphere is given as

$R_{sphere} = 1.0010 cm$

now by thermal expansion formula we know that

$L = L_o(1 + \alpha \Delta T)$

so we will have

$1.0010 = 1.0000(1 + (19\times 10^{-6})\Delta T)$

so we have

$\Delta T = 52.6 ^o C$