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

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