Solved Problem on Thermal Expansion Português English

A container with an internal volume of 1 liter at 20 °C is heated to 100 °C. Find the internal volume of this container, after heating knowing that the coefficient of linear expansion of the material is 15 × 10−6 °C−1.

Problem data
• initial internal volume:    V0 = 1 l;
• initial temperature:    ti = 20 °C;
• final temperature:    tf = 100 °C;
• coefficient of linear expansion of the container:    α = 15 × 10−6 °C−1.
Problem diagram

The problem is equivalent to a body constructed of the same material as the container and of the same internal volume V0 as the inner part of the container (figure 1-A). When the body is heated, it will expand (figure 1-B). figure 1

The heated body will reach a new volume V after expansion (figure 2-A) figure 2

The internal volume of the heated container will be the same volume V as the body (figure 2-B).

Solution

First, we convert the volume given in liters to cubic meters used in the International System of Units (S.I.)
$V=1\;\cancel{{\text{l}}} \times \frac{1\;\text{m}^{3}}{1000\;\cancel{{\text{l}}}}=\frac{1}{1000}\;\text{m}^{3}=\frac{1}{10^{3}}\;\text{m}^{3}=1 \times 10^{-3}\;\text{m}^{3}$
The problem gives us the coefficient of linear expansion of the material and for the calculation of the volume expansion, we need the coefficient of volumetric expansion
$\gamma =3\alpha \\ \gamma =3 \times 15 \times 10^{-6}\\ \gamma =45 \times 10^{-6}\;^{o}\text{C}^{-1}$
The volume of the body, made of the same material as the container, after heating will be
$\bbox[#99CCFF,10px] {\Delta V=\gamma V_{0}\Delta t}$
$V-V_{0}=\gamma V_{0}(\;t-t_{0}\;)\\ V=V_{0}+\gamma V_{0}(\;t-t_{0}\;)\\ V=V_{0}[\;1+\gamma (\;t-t_{0}\;)\;]\\ V=1 \times 10^{-3}[\;1+45 \times 10^{-6}(\;100-20\;)\;]\\ V=1 \times 10^{-3}[\;1+45 \times 10^{-6} \times 80\;]\\ V=1 \times 10^{-3}[\;1+3600 \times 10^{-6}\;]\\ V=1 \times 10^{-3}[\;1+0.0036\;]\\ V=1 \times 10^{-3} \times 1.0036$
$\bbox[#FFCCCC,10px] {V=1.004 \times 10^{-3}\;\text{m}^{3}=1.004\;\text{l}}$ 