Chapter 13, Problem 6P

To determine

Rank the increase in length from greatest to smallest.

Answer to Problem 6P

The rank of the increase in length from greatest to smallest is $\left(\text{b}\right)\text{=}\left(\text{d}\right)\text{=}\left(\text{e}\right)\text{,}\left(\text{a}\right)\text{,}\left(\text{c}\right)$.

Explanation of Solution

Write an expression for the increase in length.

$\begin{array}{c}\Delta L=L\alpha \Delta T\\ =L\alpha \left(T_f-T_i\right)\end{array}$ (I)

Here, $\Delta L$ is the increase in length, $L$ is the initial length, $\alpha$ is the coefficient of thermal expansion, $\Delta T$ is the change in temperature, $T_f$ is final temperature and $T_i$ is the initial temperature.

Conclusion:

Consider slab (a).

Substitute $90\text{cm}$ for $L$, $40°\text{C}$ for $T_f$, $20°\text{C}$ for $T_i$ and $8\times {10}^{-6}{\text{K}}^{-1}$ for $\alpha$ in equation (I) to find $\Delta L$.

$\begin{array}{c}\Delta L=\left(\left(90\text{cm}\right)\left(\frac{1\text{m}}{{10}^{-2}\text{cm}}\right)\right)\left(8\times {10}^{-6}{\text{K}}^{-1}\right)\left(\left(40°\text{C+}273.15\text{K/}°\text{C}\right)-\left(20°\text{C+}273.15\text{K/}°\text{C}\right)\right)\\ =\left(0.90\text{m}\right)\left(8\times {10}^{-6}{\text{K}}^{-1}\right)\left(20\text{K}\right)\\ =1.44\times {10}^{-4}\text{m}\end{array}$

Consider slab (b).

Substitute $90\text{cm}$ for $L$, $50°\text{C}$ for $T_f$, $20°\text{C}$ for $T_i$ and $8\times {10}^{-6}{\text{K}}^{-1}$ for $\alpha$ in equation (I) to find $\Delta L$.

$\begin{array}{c}\Delta L=\left(\left(90\text{cm}\right)\left(\frac{1\text{m}}{{10}^{-2}\text{cm}}\right)\right)\left(8\times {10}^{-6}{\text{K}}^{-1}\right)\left(\left(50°\text{C+}273.15\text{K/}°\text{C}\right)-\left(20°\text{C+}273.15\text{K/}°\text{C}\right)\right)\\ =\left(0.90\text{m}\right)\left(8\times {10}^{-6}{\text{K}}^{-1}\right)\left(30\text{K}\right)\\ =2.16\times {10}^{-4}\text{m}\end{array}$

Consider slab (c).

Substitute $60\text{cm}$ for $L$, $40°\text{C}$ for $T_f$, $20°\text{C}$ for $T_i$ and $8\times {10}^{-6}{\text{K}}^{-1}$ for $\alpha$ in equation (I) to find $\Delta L$.

$\begin{array}{c}\Delta L=\left(\left(60\text{cm}\right)\left(\frac{1\text{m}}{{10}^{-2}\text{cm}}\right)\right)\left(8\times {10}^{-6}{\text{K}}^{-1}\right)\left(\left(40°\text{C+}273.15\text{K/}°\text{C}\right)-\left(20°\text{C+}273.15\text{K/}°\text{C}\right)\right)\\ =\left(0.60\text{m}\right)\left(8\times {10}^{-6}{\text{K}}^{-1}\right)\left(20\text{K}\right)\\ =0.96\times {10}^{-4}\text{m}\end{array}$

Consider slab (d).

Substitute $90\text{cm}$ for $L$, $40°\text{C}$ for $T_f$, $20°\text{C}$ for $T_i$ and $12\times {10}^{-6}{\text{K}}^{-1}$ for $\alpha$ in equation (I) to find $\Delta L$.

$\begin{array}{c}\Delta L=\left(\left(90\text{cm}\right)\left(\frac{1\text{m}}{{10}^{-2}\text{cm}}\right)\right)\left(12\times {10}^{-6}{\text{K}}^{-1}\right)\left(\left(40°\text{C+}273.15\text{K/}°\text{C}\right)-\left(20°\text{C+}273.15\text{K/}°\text{C}\right)\right)\\ =\left(0.90\text{m}\right)\left(12\times {10}^{-6}{\text{K}}^{-1}\right)\left(20\text{K}\right)\\ =2.16\times {10}^{-4}\text{m}\end{array}$

Consider slab (e).

Substitute $60\text{cm}$ for $L$, $50°\text{C}$ for $T_f$, $20°\text{C}$ for $T_i$ and $12\times {10}^{-6}{\text{K}}^{-1}$ for $\alpha$ in equation (I) to find $\Delta L$.

$\begin{array}{c}\Delta L=\left(\left(60\text{cm}\right)\left(\frac{1\text{m}}{{10}^{-2}\text{cm}}\right)\right)\left(12\times {10}^{-6}{\text{K}}^{-1}\right)\left(\left(50°\text{C+}273.15\text{K/}°\text{C}\right)-\left(20°\text{C+}273.15\text{K/}°\text{C}\right)\right)\\ =\left(0.60\text{m}\right)\left(12\times {10}^{-6}{\text{K}}^{-1}\right)\left(30\text{K}\right)\\ =2.16\times {10}^{-4}\text{m}\end{array}$

Thus, the rank of the increase in length from greatest to smallest is $\left(\text{b}\right)\text{=}\left(\text{d}\right)\text{=}\left(\text{e}\right)\text{,}\left(\text{a}\right)\text{,}\left(\text{c}\right)$.