The temperature decreases by 95 C, so the length decreases by
L*(1.2*10^-5)*95 = 1.94*10^-2 m
= 1.94 cm
Note that I had to use 1/3 of the coefficient of volume expansion for the coefficient of linear expansion. That is becasue the iron expands in all three dimensions.
The coefficient of volume expansion for iron is 3.6 x 10-5/Co. If a 17 m long rod of iron is cooled from 100oC to 5oC, what it the new length?
3 answers
α=ΔL/L•ΔT
ΔL = α• L•ΔT=3.6•10⁻⁵•17• (5-100) =
= -0.05814 m
ΔL = α• L•ΔT=3.6•10⁻⁵•17• (5-100) =
= -0.05814 m
α=ΔL/L•ΔT
α=3.6•10⁻⁵/3
ΔL = 3.6•10⁻⁵•17• (5-100)/3 =
= -0.05814/3=0.01938 m
α=3.6•10⁻⁵/3
ΔL = 3.6•10⁻⁵•17• (5-100)/3 =
= -0.05814/3=0.01938 m