Question

A steel ball has a diameter of

Answer 1

The required increase in temperature is: $157^{\circ}C$

Explanation:

When the temperature increases, the area of an object increases according to the equation

$\Delta A = 2\alpha A_0 \Delta T$

where

$A_0$ is the initial area

$\alpha$ is the coefficient of linear expansion

$\Delta A$ is the change in area

$\Delta T$ is the change in temperature

For the steel ball in this problem, we have:

$d=0.125 m$, so the radius is

$r=\frac{0.125}{2}=0.0625 m$

Therefore, the surface area is

$A=4\pi r^2 = 4\pi (0.0625)^2=0.0491 m^2$

The coefficient of linear expansion for steel is

$\alpha = 13\cdot 10^{-6} ^{\circ}C^{-1}$

And the change in area of the ball is

$\Delta A = 0.0002 m^2$

Therefore, the required increase in temperature is:

$\Delta T = \frac{\Delta A}{2\alpha A_0}=\frac{0.0002}{2(13\cdot 10^{-6})(0.0491)}=157^{\circ}C$

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