r2 = r1 + a*(T-To)r1 = 4.42 cm.
4.40 + 1.3*10^-4(T-20)4.40 = 4.42
4.40 + 5.72*10^-4(T-20) = 4.42
5.72*10^-4(T-20) = 4.42-4.40 = 0.02
5.72*10^-4T - 114.4*10^-4 = 0.02
5.72*10^-4T = 0.02 + 0.01144 = 0.03144
T = 314.4*10^-4/5.72*10^-4 = 54.97o
A pair of eyeglass frames made of plastic with a coefficient of linear expansion of
1.30x10-4
(Cº
)
-1
have circular lens holes of diameter 4.40 cm at room temp (20º
C).
To what final temperature must the frames be heated in order to insert lenses
4.42 cm in diameter? (Hint: treat this as an area expansion problem)
1 answer
0 answers