Question

A brass plug is to be placed in a ring made of iron. At 20°C, the diameter of the plug is 8.740 cm and that of the inside of the ring is 8.730 cm.

(a) They must both be brought to what common temperature in order to fit?

(b) Repeat part (a) if the plug were iron and the ring brass.

Answer #1

L_{ro} = initial diameter of ring = 8.73 cm

_{r}
= coefficient of thermal expansion for iron = 12 x
10^{-6}

L_{po} = initial diameter of ring = 8.74 cm

_{b}
= coefficient of thermal expansion for brass = 19 x
10^{-6}

length of ring = length of brass

L_{ro} + L_{ro}_{r}
(T - 20) = L_{ro} + L_{bo}_{b}
(T - 20)

8.73 + (8.73) (12 x 10^{-6}) (T - 20) = 8.74 + (8.74)
(19 x 10^{-6}) (T - 20)

T = - 143.13

b)

L_{ro} = initial diameter of plug = 8.73 cm

_{r}
= coefficient of thermal expansion for iron = 12 x
10^{-6}

L_{po} = initial diameter of plug = 8.74 cm

_{b}
= coefficient of thermal expansion for brass = 19 x
10^{-6}

length of ring = length of brass

L_{ro} + L_{ro}_{r}
(T - 20) = L_{ro} + L_{bo}_{b}
(T - 20)

8.74 + (8.74) (12 x 10^{-6}) (T - 20) = 8.73 + (8.73)
(19 x 10^{-6}) (T - 20)

T = 183.86

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