The coefficient of linear expansion of copper is $17 \times 10 ^ { - 6 } K - 1$ A sheet of copper has a round hole with a radius of 3.0 m cut out of it. If the sheet is heated and undergoes a change in temperature of 80 K, what is the change in the radius of the hole?
A) It decreases by 4.1 mm
B) It decreases by 8.2 mm
C) It does not change.
D) It increases by 4.1 mm
E) It increases by 8.2 mm

Question

The coefficient of linear expansion of copper is $17 \times 10 ^ { - 6 } K - 1$ A sheet of copper has a round hole with a radius of  3.0 m cut out of it. If the sheet is heated and undergoes a change in temperature of  80 K, what is the change in the radius of the hole?
A) It decreases by  4.1 mm
B) It decreases by  8.2 mm
C) It does not change.
D) It increases by  4.1 mm
E) It increases by  8.2 mm

Quizplus · Accepted Answer

The radius of the hole increases due to thermal expansion. The change in radius $\Delta r$ can be calculated using the formula $\Delta r = \alpha \cdot r \cdot \Delta T$, where $\alpha = 17 \times 10^{-6} K^{-1}$, $r = 3.0 m$, and $\Delta T = 80 K$. Substituting these values, $\Delta r = 17 \times 10^{-6} \cdot 3.0 \cdot 80 = 4.08 \times 10^{-3} m$ or 4.1 mm.

The Coefficient of Linear Expansion of Copper Is 17×10−6K−117 \times 10 ^ { - 6 } K - 117×10−6K−1

The Coefficient of Linear Expansion of Copper Is $17 \times 10 ^ { - 6 } K - 1$