Using equipartition of energy, the specific heat (in J kg$$^{-1}$$ K$$^{-1}$$) of Aluminium at high temperature can be estimated to be (atomic weight of Aluminium = 27)

According to the equipartition theorem of energy, each quadratic degree of freedom contributes an average energy of $$\dfrac12 k_B T$$ to a system in thermal equilibrium, where $$k_B$$ is the Boltzmann constant and $$T$$ is the absolute temperature.

An atom bound inside a crystalline solid behaves like a three-dimensional harmonic oscillator. Such an oscillator possesses

• three kinetic energy terms (one for motion along each axis), and
• three potential energy terms (one for the restoring force along each axis).

Thus the number of quadratic terms per atom is

$$6 = 3 \text{(kinetic)} + 3 \text{(potential)}.$$

Using equipartition, the average energy of one atom is therefore

$$\bigl(6\bigr)\left(\dfrac12 k_B T\right)=3 k_B T.$$

For one mole we have Avogadro’s number $$N_A$$ of atoms, so the molar internal energy becomes

$$U_{\text{molar}} = N_A \left(3 k_B T\right)=3 N_A k_B T.$$

But the universal gas constant is defined as $$R = N_A k_B,$$ so we can write

$$U_{\text{molar}} = 3 R T.$$

The molar heat capacity at constant volume is the temperature derivative of this energy:

$$C_V^{\text{molar}} = \dfrac{dU_{\text{molar}}}{dT}=3R.$$

At high temperatures the difference between $$C_P$$ and $$C_V$$ for a solid is negligible, so we may take the molar specific heat simply as

$$C^{\text{molar}} = 3R.$$

The numerical value of the gas constant is

$$R = 8.314\ \text{J mol}^{-1}\ \text{K}^{-1}.$$

Hence

$$C^{\text{molar}} = 3 \times 8.314 = 24.942\ \text{J mol}^{-1}\ \text{K}^{-1}.$$

To convert this molar specific heat to a mass specific heat (J kg$$^{-1}$$ K$$^{-1}$$) we divide by the molar mass (atomic weight expressed in kg per mole). Aluminium has an atomic weight of 27 g mol$$^{-1}$$, which is

$$M = 27\ \text{g mol}^{-1} = 0.027\ \text{kg mol}^{-1}.$$

Therefore the specific heat per kilogram is

$$ C^{\text{mass}} = \dfrac{C^{\text{molar}}}{M} = \dfrac{24.942\ \text{J mol}^{-1}\ \text{K}^{-1}} {0.027\ \text{kg mol}^{-1}} = 924.9\ \text{J kg}^{-1}\ \text{K}^{-1}. $$

Rounding to a convenient integer gives approximately $$925\ \text{J kg}^{-1}\ \text{K}^{-1}.$$

Hence, the correct answer is Option D.