Question

Uma barra de ferro tem comprimento 30 m a 0°C. Sabendo-se que o coeficiente de dilatação linear do ferro é igual a 12 x 10 - 6 °C-1, calcular:

a)     O comprimento final da barra a 20°C

b)     O comprimento final da barra a -30°C​

Answer 1

Para resolver sua questão, temos a seguinte fórmula:

  $\Large\displaystyle\begin{gathered} \sf \Delta L= L_0\cdot \alpha \cdot \Delta \theta\end{gathered}$

• Sendo que:

$\large\displaystyle\begin{gathered} \sf \Delta L\implies Variac_{\!\!\!,}\tilde{a}o\ do\ comprimento\ [m]\end{gathered}$

$\large\displaystyle\begin{gathered} \sf L_0\implies Comprimento\ inicial\ [m]\end{gathered}$

$\large\displaystyle\begin{gathered} \sf \alpha\implies Coeficiente\ de\ dilatac_{\!\!\!,}\tilde{a}o\ linear\ [^{\circ}C^{-1}]\end{gathered}$

$\large\displaystyle\begin{gathered} \sf \Delta \theta\implies Variac_{\!\!\!,}\tilde{a}o\ da \ temperatura\ [^{\circ}C]\end{gathered}$

Importante lembrar que quando temos variação de algo, quer dizer que o final menos inicial é igual a variação. Ou seja, $\large\displaystyle\begin{gathered} \sf \Delta L=L-L_0\end{gathered}$ e $\large\displaystyle\begin{gathered} \sf \Delta \theta=\theta-\theta_0\end{gathered}$. Com isso, vamos resolver o item a:

$\Large\displaystyle\begin{gathered} \sf \Delta L= L_0\cdot \alpha \cdot \Delta \theta\end{gathered}$

$\Large\displaystyle\begin{gathered} \sf L-L_0= L_0\cdot \alpha \cdot \Delta \theta\end{gathered}$

$\Large\displaystyle\begin{gathered} \sf L= L_0\cdot \left(1+ \alpha \cdot \Delta \theta\right)\end{gathered}$

$\Large\displaystyle\begin{gathered} \sf L= 30\cdot \left(1+ 12\cdot 10^{-6} \cdot 20\right)\end{gathered}$

$\Large\displaystyle\begin{gathered} \sf L= 30+ 7200\cdot 10^{-6} \end{gathered}$

$\Large\displaystyle\begin{gathered} \sf L= 30+ 0,0072\cdot 10^6\cdot 10^{-6} \end{gathered}$

$\Large\displaystyle\begin{gathered} \sf L= 30+ 0,0072 \end{gathered}$

$\Large\displaystyle\begin{gathered} \boxed{\sf L= 30,0072\ m }\end{gathered}$

Item b:

$\Large\displaystyle\begin{gathered} \sf \Delta L= L_0\cdot \alpha \cdot \Delta \theta\end{gathered}$

$\Large\displaystyle\begin{gathered} \sf L-L_0= L_0\cdot \alpha \cdot \Delta \theta\end{gathered}$

$\Large\displaystyle\begin{gathered} \sf L= L_0\cdot \left(1+ \alpha \cdot \Delta \theta\right)\end{gathered}$

$\Large\displaystyle\begin{gathered} \sf L= 30\cdot \left(1+ 12\cdot 10^{-6} \cdot (-30)\right)\end{gathered}$

$\Large\displaystyle\begin{gathered} \sf L= 30- 10800\cdot 10^{-6} \end{gathered}$

$\Large\displaystyle\begin{gathered} \sf L= 30- 0,0108\cdot 10^6\cdot 10^{-6} \end{gathered}$

$\Large\displaystyle\begin{gathered} \sf L= 30- 0,0108\end{gathered}$

$\Large\displaystyle\begin{gathered} \boxed{\sf L= 29,9892\ m}\end{gathered}$

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