Answer:
[tex]\Delta V = 1.440\times 10^{-6}\,m^{3}[/tex]
Explanation:
El coeficiente de expansión volumétrica se calcula a partir de la siguiente ecuación diferencial parcial (The volumetric expansion coefficient is computed by means of the following partial differential equation):
[tex]\alpha = \frac{1}{V} \cdot \left(\frac{\partial V}{\partial T} \right)[/tex]
Se integra la fórmula a continuación (The formula is integrated herein):
[tex]\alpha\, dT = \frac{dV}{V}[/tex]
Supóngase que el coeficiente es constante (Let suppose that coefficient is constant):
[tex]\alpha \int\limits^{T_{f}}_{T_{o}}\,dT = \int\limits^{V_{f}}_{V_{o}}\, \frac{dV}{V}[/tex]
[tex]\alpha \cdot (T_{f}-T_{o}) = \ln \frac{V_{f}}{V_{o}}[/tex]
El volumen final es (The final volume is):
[tex]V_{f} = V_{o}\cdot e^{\alpha \cdot (T_{f}-T_{o})}[/tex]
El coeficiente de expansión volumétrica del acero es [tex]12\times 10^{-6}\,^{\circ}C^{-1}[/tex] (The volumetric expansion coefficient of steel is [tex]12\times 10^{-6}\,^{\circ}C^{-1}[/tex]):
[tex]V_{f} = (0.004\,m^{3})\cdot e^{(12\times 10^{-6}\,^{\circ}C^{-1})\cdot (50^{\circ}C-20^{\circ}C)}[/tex]
[tex]V_{f} \approx 4.001\times 10^{-3}\,m^{3}[/tex]
Finalmente, la dilatación experimentada por el balín es (Lastly, the dillatation experimented by the pellet is):
[tex]\Delta V = 4.001\times 10^{-3}\,m^{3} - 4.000\times 10^{-3}\,m^{3}[/tex]
[tex]\Delta V = 1.440\times 10^{-6}\,m^{3}[/tex]
