Young’s modulus for copper is measured by stretching a copper wire to be about $\mathbf{1}\mathbf{.}\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{11}}\mathbf{N}\mathbf{/}{\mathbf{m}}^{\mathbf{2}}$. The density of copper is about $\mathbf{9}\mathbf{}\mathbf{g}\mathbf{/}{\mathbf{cm}}^{\mathbf{3}}$, and the mass of a mole is .Starting from a very low temperature, use these data to estimate roughly the temperature T at which we expect the specific heat for copper to approach 3 kB . Compare your estimate with the data shown on a graph in this chapter.

The given data can be listed below,

• The young modulus of copper wire is, $\mathrm{Y}=1.2\times {10}^{11}\mathrm{N}/{\mathrm{m}}^2$
• The density of copper wire is, $\left(\frac{1\mathrm{kg}}{1000\mathrm{g}}\right)\left(\frac{1{\mathrm{cm}}^3}{{10}^{-6}{\mathrm{m}}^3}\right)=9\times {10}^3\mathrm{kg}/{\mathrm{m}}^3$.
• The mass of one mole of copper is, ${\mathrm{m}}_{\mathrm{cu}}=\left(\frac{63.5\mathrm{g}}{6.02\times 10}\right)\left(\frac{{10}^{-3}\mathrm{kg}}{1\mathrm{g}}\right)=1.05\times {10}^{-25}\mathrm{kg}$.
• The specific heat of the copper wire is, $\mathrm{C}=3{\mathrm{k}}_{\mathrm{B}}=4.14\times {10}^{-23}\mathrm{J}/\mathrm{K}$

The heat capacity of a sample of a substance divided by the mass of the sample, also known as massic heat capacity, is the specific heat capacity of the substance.

The spacing between copper atoms is given by,

$d={\left(\frac{m_{cu}}{\rho }\right)}^{1/3}$

Here,${\mathrm{m}}_{\mathrm{cu}}$is the mass of copper and $\rho$is he density of copper.

Substitute all the values in the above,

$\mathrm{d}={\left(\frac{1.05\times {10}^{-25}\mathrm{kg}}{9\times {10}^3\mathrm{kg}/{\mathrm{m}}^3}\right)}^{1/3}$

$=2.27\times {10}^{-10}\mathrm{m}$

Stiffness of the copper wire is given by,

$k_s=Yd$

Here, Y is the young modulus andd is the distance between copper atoms.

Substitute values in the above,

$$\begin{gathered} {\mathrm{k}}_{\mathrm{s}}=\left(1.2\times {10}^{-11}\mathrm{N}/{\mathrm{m}}^2\right)\left(2.27\times {10}^{-10}\mathrm{m}\right) \\ =27.24\mathrm{N}/\mathrm{m} \end{gathered}$$

The energy transferred to copper is given by,

$$\begin{gathered} ∆E=\sout{h}\omega \\ =\sout{h}\sqrt{\frac{k_s}{m_{cu}}} \end{gathered}$$

Here, $\sout{h}$is the Planck’s constant, $k_s$is the stiffness of the copper wire and is the mass of the copper.

Substitute all the values in the above,

$$\begin{gathered} ∆\mathrm{E}=\left(1.05\times {10}^{-34}\mathrm{J}.\mathrm{S}\right)\sqrt{\frac{27.24\mathrm{N}/\mathrm{m}}{1.05\times {10}^{-25}\mathrm{kg}}} \\ =1.69\times {10}^{-21}\mathrm{J} \end{gathered}$$

The temperature of the copper wire is given by,

$T=\frac{∆E}{C}$

Here, $∆E$is the energy transferred to copper wire, and C is the specific heat capacity.

Substitute all the values in the above,

$$\begin{gathered} \mathrm{T}=\frac{1.69\times {10}^{-21}\mathrm{J}}{4.14\times {10}^{23}\mathrm{J}/\mathrm{K}} \\ =40.82\mathrm{K} \end{gathered}$$

From graph it is clear that the temperature equal to 400 K at specific heat capacity is $3k_B$ and the calculated value is less than the value in the graph.