Missing question:
"What is the gap between the slabs when the temperature is 0∘F?"
Solution)
The linear expansion of a solid material that undergoes a temperature change of [tex]\Delta T[/tex] is given by
[tex]\Delta L = \alpha _L L \Delta T [/tex]
where [tex]\alpha _L[/tex] is the coefficient of linear expansion of the material, L the length and [tex]\Delta L[/tex] the linear expansion.
For concrete, [tex]\alpha _L = 12 \cdot 10^{-6} K^{-1}[/tex]. We also have the length of the slab, [tex]L=315 cm=3.15 m[/tex].
To find the temperature variation, we must convert first the temperatures in Kelvin:
[tex]T_f = 115 F=319.3 K[/tex]
[tex]T_i = 0 F=255.4 K[/tex]
Therefore, we can find the linear expansion:
[tex]\Delta L = (12\cdot 10^{-6} K^{-1})(319.3 K-255.4 K)(3.15 m)=0.0024 m=2.4 mm[/tex]
This is the variation in length between the two temperatures T=115 F and T=0 F. Therefore, since the slabs just touch at 115 F, their distance is 0, while at T=0 F their gap will be exactly 2.4 mm.
