Volume change: Given V0 = 0.75 m^3, beta = 6.0 × 10^-6 /°C, delta T = 25 °C, what is delta V?

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Multiple Choice

Volume change: Given V0 = 0.75 m^3, beta = 6.0 × 10^-6 /°C, delta T = 25 °C, what is delta V?

Explanation:
Volumetric expansion: when a material heats up, its volume increases in proportion to a coefficient called beta, the volumetric thermal expansion coefficient. For small temperature changes, the change in volume is delta V = V0 * beta * delta T. Plug in the numbers: V0 = 0.75 m^3, beta = 6.0 × 10^-6 /°C, delta T = 25 °C. Multiply beta by delta T: 6.0 × 10^-6 × 25 = 1.5 × 10^-4. Then delta V = 0.75 × 1.5 × 10^-4 = 1.125 × 10^-4 m^3, which is 0.0001125 m^3. So the volume increases by about 1.125 × 10^-4 cubic meters (roughly 0.1125 liters). The small result makes sense because the expansion coefficient is a small number and the temperature change, while moderate, is not huge. If you saw a much larger change, it would come from using a larger delta T or a larger beta, which would push delta V into the 0.0002–0.001 range or higher.

Volumetric expansion: when a material heats up, its volume increases in proportion to a coefficient called beta, the volumetric thermal expansion coefficient. For small temperature changes, the change in volume is delta V = V0 * beta * delta T.

Plug in the numbers: V0 = 0.75 m^3, beta = 6.0 × 10^-6 /°C, delta T = 25 °C. Multiply beta by delta T: 6.0 × 10^-6 × 25 = 1.5 × 10^-4. Then delta V = 0.75 × 1.5 × 10^-4 = 1.125 × 10^-4 m^3, which is 0.0001125 m^3. So the volume increases by about 1.125 × 10^-4 cubic meters (roughly 0.1125 liters).

The small result makes sense because the expansion coefficient is a small number and the temperature change, while moderate, is not huge. If you saw a much larger change, it would come from using a larger delta T or a larger beta, which would push delta V into the 0.0002–0.001 range or higher.

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