A bore made in an aluminium block at 34ºC is 3.48cm$^3$. What is the new bore when the temperature was raised to 340ºC [α$_a$ = 24 x 10$^{-6}$K$^{-1}$]

Analysis

This is a problem involving Cubical Expansion (or Volume Expansion) of a solid. Even though it is a bore (a hole) in a solid block, the hole expands exactly as if it were made of the same material as the block.

Concept

The formula for the change in volume is:

Where:

• $V_0$ = Initial volume ($3.48 \text{ cm}^3$)
• $\gamma$ = Coefficient of cubical expansion ($\gamma = 3\alpha$)
• $\alpha$ = Linear expansivity ($24 \times 10^{-6} \text{ K}^{-1}$)
• $\Delta\theta$ = Change in temperature ($T_2 - T_1$)
• $V_{new} = V_0 + \Delta V$

Step-by-Step Solution

1. Identify the given values:

• $V_0 = 3.48 \text{ cm}^3$
• $\alpha = 24 \times 10^{-6} \text{ K}^{-1}$
• $\gamma = 3 \times \alpha = 3 \times 24 \times 10^{-6} = 72 \times 10^{-6} \text{ K}^{-1}$
• $\Delta\theta = 340^\circ\text{C} - 34^\circ\text{C} = 306^\circ\text{C}$ (or $306\text{ K}$)

2. Calculate the change in volume ($\Delta V$):

3. Calculate the new volume ($V_{new}$):

Rounding to two decimal places, we get 3.56 cm$^3$.

Conclusion

Comparing our result to the options provided, the calculated value matches option C.

Exam Tip: Always remember that for holes in solids, the hole expands as if it were filled with the material of the solid. Never make the mistake of thinking the hole shrinks! Also, ensure you multiply the linear expansivity ($\alpha$) by 3 to get the cubical expansivity ($\gamma$) before calculating volume changes.