p19_012.pdf

12. (a) The coecient of linear expansion α for the alloy is

α =∆ L/L ∆ T =

10 . 000cm

(10 . 01cm)(100 ◦ C

−

20 . 000 ◦ C) =1 . 88 × 10 − 5 / C ◦ .

−

Thus, from 100 ◦ Cto0 ◦ Cwehave

∆ L = Lα ∆ T =(10 . 015cm) 1 . 88

10 − 5 / C ◦ (0 ◦ C

−

100 ◦ C) =

−

1 . 88

10 − 2 cm .

0 . 0188cm = 9 . 996cm .

(b) Let the temperature be T x .Thenfrom20 ◦ Cto T x wehave

−

∆ L =10 . 009cm

−

10 . 000cm = αL ∆ T =(1 . 88

10 − 5 / C ◦ )(10 . 000cm)∆ T,

giving ∆ T =48C ◦ .Thus, T x =20 ◦ C+48C ◦ =68 ◦ C .

10 . 015cm

The length at 0 ◦ C is therefore L = L +∆ L =10 . 015cm