Chapter 29, Problem 34GP

To determine

The latent heat of fusion for the given solid.

Answer to Problem 34GP

  $2.8\times {10}^{4}J/kg$ .

Explanation of Solution

Given:

  

From the given figure,

• An atom inside cube is bonded to 6 nearest neighbours.
• The number of bonds per atom is 3.

Formula used:

It is known that: ${\text{L}}_{\text{fusion}}=\left(\frac{\text{number}\text{of}\text{bonds}}{\text{atoms}}\right)\left(\frac{\text{number}\text{of}\text{bonds}}{\text{kg}}\right){\text{E}}_{\text{bond}}$

Where,

  ${\text{L}}_{\text{fusion}}$ is the latent heat for fusion.

Calculation:

Finding the heat of fusion for argon from the energy required to break the bond: ${\text{L}}_{\text{fusion}}=\left(\frac{\text{number}\text{of}\text{bonds}}{\text{atoms}}\right)\left(\frac{\text{number}\text{of}\text{bonds}}{\text{kg}}\right){\text{E}}_{\text{bond}}$

  $\therefore {\text{L}}_{\text{fusion}}=\left(3\right)\left[\frac{6.02\times {10}^{23}\text{atoms/mol}}{39.95\times {10}^{-3}\text{kg/mol}}\right]\times \left(3.9\times {10}^{-3}\text{eV}\right)\left(1.60\times {10}^{-19}\text{J/eV}\right)$

  $\therefore {\text{L}}_{\text{fusion}}=2.8\times {10}^4\text{J/kg}$

Conclusion: The latent heat of fusion for the given solid is $2.8\times {10}^4\text{J/kg}$ .