Chapter 13, Problem 86A

To determine

Increase in sphere volume.

Answer to Problem 86A

Increase in volume will be $7.6\times {10}^{-2}{\text{cm}}^3$

Explanation of Solution

Given:

Volume of sphere, $\text{= 1}\text{.78}{\text{cm}}^3$

Initial temperature $\text{= 11}°\text{C}$

Final temperature $\text{= 580}°\text{C}$

Volume coefficient of expansion of aluminum, $\beta \text{ = 75}\times {\text{10}}^{-6}\text{per}°\text{C}$

Formula used:

Change in volume:

  $\text{Δ}V=\beta V\text{Δ}T$

Where, $\text{Δ}V$ is change in volume, $\beta$ is coefficient of volume expansion, $V$ is the volume, $\Delta T$ is the change in temperature.

Calculation:

Substituting the values:

  $\text{Δ}V=\beta V\text{Δ}T$

  $\Delta V=75\times {10}^{-6}\text{per}°\text{C}\times 1.78{\text{cm}}^3\times \left(580-11\right)°\text{C}$

  $\Delta V=\text{7}\text{.6}\times {\text{10}}^{-2}{\text{cm}}^3$

Conclusion:

Aluminum sphere volume will increase by $7.6\times {10}^{-2}{\text{cm}}^3$ .