Question

Using equipartition of energy, the specific heat $\left(Jk{g}^{-1}{K}^{-1}\right)$ of aluminium at high temperature can be estimated to be (atomic weight of aluminium is $27$).
(A) $25$
(B) $1850$
(C) $410$
(D) $923$

Answer 1

Hint:
The theory of the equipartition of energy helps to find the solution for this problem and also by using the kinetic energy formula and the heat capacity formula the specific heat of aluminium at the high temperature can be determined.
Useful formula:
By law of Equipartition of energy,
${\displaystyle \frac{3}{2}}{K}_{B}T={\displaystyle \frac{1}{2}}mCT$
Where, $K_B$ is the Boltzmann constant, $T$ is the temperature, $m$ is the atomic mass of the aluminium and $C$ is the specific heat of the aluminium.

 Complete step-by-step solution:
Given that,
The atomic weight of aluminium is $27g$,
By law of Equipartition of energy,
${\displaystyle \frac{3}{2}}{K}_{B}T={\displaystyle \frac{1}{2}}mCT..................\left(1\right)$
By cancelling the temperature value on both side in the above equation, then the above equation is written as,
${\displaystyle \frac{3}{2}}{K}_B={\displaystyle \frac{1}{2}}mC$
By cancelling the denominator on both sides, then the above equation is written as,
 $3K_B=mC$
By keeping the specific heat of the aluminium in one side and the other terms in other side, then the above equation is written as,
$C={\displaystyle \frac{3\times K_B}{m}}$
Here to find the specific heat capacity the Avogadro number $\left(N_A\right)$ should be multiplied with RHS, then the above equation is written as,
$C={\displaystyle \frac{3\times K_B}{m}}\times N_A.................\left(2\right)$
On substituting the known values of Boltzmann constant, Avogadro number and the atomic mass of aluminium in the above equation (2), then the above equation is written as,
$C={\displaystyle \frac{3\times 1.38\times {10}^{-23}\times 6.02\times {10}^{23}}{27\times {10}^{-3}}}$
In numerator, the terms ${10}^{-23}$ and the term ${10}^{23}$ gets cancelled, then the above equation is written as,
$C={\displaystyle \frac{3\times 1.38\times 6.02}{27\times {10}^{-3}}}$
On multiplying the terms in numerator,
$C={\displaystyle \frac{24.9228}{27\times {10}^{-3}}}$
By taking the term ${10}^{-3}$ from denominator to the numerator, then the above equation is written as,
$C={\displaystyle \frac{24.9228\times {10}^3}{27}}$
On multiplying the terms in numerator, then
$C={\displaystyle \frac{24922.8}{27}}$
Now dividing the terms, then the above equation is written as,
$C=923.066Jk{g}^{-1}{K}^{-1}$
Thus, the above equation shows the specific heat of the aluminium.
Hence, the option (D) is the correct answer.

Note:-
The reason for multiplying the Avogadro number in the solution, the mass of aluminium is given as atomic mass so the Avogadro number is multiplied. While substituting the mass value it must be in terms of kilogram, so the term ${10}^{-3}$ is multiplied with the atomic mass value.