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,
[tex]\alpha=\frac {\Delta r}{r\times \Delta T}[/tex]
Also,
[tex]\Delta T=T_{final}-T_{Initial}[/tex]
So,
[tex]19\times 10^{-6}=\frac {0.0021}{1.3290\times (T_{final}-23.0)}[/tex]
Solving for final temperature as:
[tex](T_{final}-23.0)=\frac {0.0021}{1.3290\times 9\times 10^{-6}}[/tex]
Final temperature = 106.165 °C
