Respuesta :
The required increase in temperature is: [tex]157^{\circ}C[/tex]
Explanation:
When the temperature increases, the area of an object increases according to the equation
[tex]\Delta A = 2\alpha A_0 \Delta T[/tex]
where
[tex]A_0[/tex] is the initial area
[tex]\alpha[/tex] is the coefficient of linear expansion
[tex]\Delta A[/tex] is the change in area
[tex]\Delta T[/tex] is the change in temperature
For the steel ball in this problem, we have:
[tex]d=0.125 m[/tex], so the radius is
[tex]r=\frac{0.125}{2}=0.0625 m[/tex]
Therefore, the surface area is
[tex]A=4\pi r^2 = 4\pi (0.0625)^2=0.0491 m^2[/tex]
The coefficient of linear expansion for steel is
[tex]\alpha = 13\cdot 10^{-6} ^{\circ}C^{-1}[/tex]
And the change in area of the ball is
[tex]\Delta A = 0.0002 m^2[/tex]
Therefore, the required increase in temperature is:
[tex]\Delta T = \frac{\Delta A}{2\alpha A_0}=\frac{0.0002}{2(13\cdot 10^{-6})(0.0491)}=157^{\circ}C[/tex]
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Answer: 133 ºC
ΔL = α*ΔT*L
0.000200 m = 12e-6 /ºC * ΔT * 0.125m
ΔT = 133 ºC