Question

A silver sphere with radius 1.3611 cm at 23.0°C must slip through a brass ring that has an internal radius of 1.3590 cm at the same temperature. To what temperature must the ring be heated so that the sphere, still at 23.0°C, can just slip through?

Answer 1

Answer:

The temperature must the ring be heated so that the sphere can just slip through is 106.165 °C.

Explanation:

For brass:

Radius = 1.3590 cm

Initial temperature = 23.0 °C

The sphere of radius 1.3611 cm must have to slip through the brass. Thus, on heating the brass must have to attain radius of 1.3611 cm

So,

Δ r = 1.3611 cm - 1.3590 cm = 0.0021 cm

The linear thermal expansion coefficient of a metal is the ratio of the change in the length per 1 degree temperature to its length.

Thermal expansion for brass = 19×10⁻⁶ °C⁻¹

Thus,

$\alpha=\frac {\Delta r}{r\times \Delta T}$

Also,

$\Delta T=T_{final}-T_{Initial}$

So,

$19\times 10^{-6}=\frac {0.0021}{1.3290\times (T_{final}-23.0)}$

Solving for final temperature as:

$(T_{final}-23.0)=\frac {0.0021}{1.3290\times 9\times 10^{-6}}$

Final temperature = 106.165 °C