Solved Problem on Thermal Expansion

Two metallic blocks A and B, at 0 °C, have equal volumes to 250.75 cm³ and 250 cm³, respectively. The averages coefficients of linear expansion are, respectively, 2×10⁻⁵ °C⁻¹ and 3×10⁻⁵ °C⁻¹. Determine:
a) The temperature in which the blocks have equal volumes;
b) What is the volume of the blocks in the temperature calculated in item (a).

Problem data:

Initial volume of block A: V_0A = 250.75 cm³;
Initial volume of block B: V_0B = 250 cm³;
Coefficient of linear expansion of block A: α_A = 2× 10⁻⁵ °C⁻¹;
Coefficient of linear expansion of block B: α_B = 3× 10⁻⁵ °C⁻¹;
Initial system temperature: t₀ = 0 °C.

Problem diagram:

Figure 1

Solution

a) The problem gives the coefficients of linear expansion of the blocks, and we need the coefficient of volumetric expansion

\[ \bbox[#99CCFF,10px] {\gamma =3\alpha} \]

For block A:

\[ \begin{gather} \gamma_{A}=3\alpha_{A}\\ \gamma_{A}=3\times 2\times 10^{-5}\\ \gamma_{A}=6\times 10^{-5}\;°\text{C}^{-1} \end{gather} \]

For block B:

\[ \begin{gather} \gamma _{B}=3\alpha_{B}\\ \gamma_{B}=3\times 6\times 10^{-5}\\ \gamma_{B}=9\times 10^{-5}\;°\text{C}^{-1} \end{gather} \]

The final volume is given by

\[ \begin{gather} \bbox[#99CCFF,10px] {V=V_{0}(1+\gamma \Delta t)} \tag{I} \end{gather} \]

Writing the expression (I) for the two blocks, we have

\[ \begin{gather} V_{A}=V_{0A}[1+\gamma_{A}(t-t_{0})]\\ V_{A}=250.75\times [1+6\times 10^{-5}\times (t-0)]\\ V_{A}=250.75\times [1+6\times 10^{-5}t] \tag{II} \end{gather} \]

\[ \begin{gather} V_{B}=V_{0B}[1+\gamma_{B}(t-t_{0})]\\ V_{B}=250\times [1+9\times 10^{-5}\times (t-0)]\\ V_{B}=250\times [1+9\times 10^{-5}t] \tag{III} \end{gather} \]

With the condition that volumes are equal, equating expressions (II) and (III)

\[ \begin{gather} V_{A}=V_{B}\\ 250.75.[1+6\times 10^{-5}t]=250\times [1+9\times 10^{-5}t]\\ 250.75+250.75\times 6\times 10^{-5}t=250+250\times 9\times 10^{-5}t\\ 250.75+1504.5\times 10^{-5}t=250+2250\times 10^{-5}t\\ 2250\times 10^{-5}t-1504.5\times 10^{-5}t=250.75-250\\ 745.5\times 10^{-5}t=0.75\\ t=\frac{0.75}{745.5\times 10^{-5}}\\ t=\frac{0.75\times 10^{5}}{745.5}\\ t=\frac{75000}{745.5} \end{gather} \]

\[ \bbox[#FFCCCC,10px] {t\approx 100.6\;°\text{C}} \]

b) Substituting the result of item (a) into the expression (II), we have the volume of the blocks

\[ \begin{gather} V_{A}=250.75\times [1+6\times 10^{-5}\times 100.6]\\ V_{A}=250.75\times [1+603.6\times 10^{-5}]\\ V_{A}=250.75\times [1+0.006036]\\ V_{A}=250.75\times [1.006036] \end{gather} \]

\[ \bbox[#FFCCCC,10px] {V_{A}=V_{B}\approx 252.3\;\text{cm}^{3}} \]

Note: We would obtain the same result if we had substituted the temperature into the expression (III)

\[ \begin{gather} V_{B}=250\times [1+9\times 10^{-5}\times 100.6]\\ V_{B}=250\times [1+905.4\times 10^{-5}]\\ V_{B}=250\times [1+0.009054]\\ V_{B}=250\times [1.009054]\\ V_{B}=V_{A}\approx 252.3\;\text{cm}^{3} \end{gather} \]

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