using euipartion of energy, the specific heat `("in" jkg^(-1)K^(-1)`of aluminium at room temperature can be estimated to be (atomic weigh of aluminium=27)

Options

A. `410`
B. `925`
C. `25`
D. `1850`

Explanation

To estimate the specific heat of aluminum at room temperature using the equipartition of energy, we can follow these steps:

### Step 1: Understand the Equipartition of Energy
According to the equipartition theorem, the total energy of a system in thermal equilibrium is shared equally among all degrees of freedom. For a monatomic ideal gas, each degree of freedom contributes \(\frac{1}{2} kT\) to the energy, where \(k\) is the Boltzmann constant and \(T\) is the temperature in Kelvin.

### Step 2: Write the Total Energy Equation
The total energy \(E\) can be expressed as:
\[ E = ms \Delta T \]
where:
- \(m\) is the mass of the substance,
- \(s\) is the specific heat capacity,
- \(\Delta T\) is the change in temperature.

### Step 3: Relate Total Energy to Degrees of Freedom
For a monatomic gas, the total kinetic energy is given by:
\[ E = \frac{3}{2} NkT \]
where \(N\) is the number of molecules. Since \(N = \frac{m}{M} \cdot N_A\) (where \(M\) is the molar mass and \(N_A\) is Avogadro's number), we can rewrite this as:
\[ E = \frac{3}{2} \left(\frac{m}{M} N_A\right) kT \]

### Step 4: Equate the Two Expressions for Energy
Setting the two expressions for energy equal gives:
\[ ms \Delta T = \frac{3}{2} \left(\frac{m}{M} N_A\right) kT \]
Assuming \(\Delta T\) is from 0 to \(T\), we can simplify this to:
\[ ms = \frac{3}{2} \frac{N_A kT}{M} \]

### Step 5: Solve for Specific Heat \(s\)
Rearranging the equation to solve for \(s\):
\[ s = \frac{3N_A k}{M} \]

### Step 6: Substitute Known Values
We know:
- \(k = 1.38 \times 10^{-23} \, \text{J/K}\)
- \(M = 27 \, \text{g/mol} = 27 \times 10^{-3} \, \text{kg/mol}\)
- \(N_A = 6.02 \times 10^{23} \, \text{mol}^{-1}\)

Substituting these values into the equation for \(s\):
\[ s = \frac{3 \times (6.02 \times 10^{23}) \times (1.38 \times 10^{-23})}{27 \times 10^{-3}} \]

### Step 7: Calculate the Specific Heat
Calculating the above expression:
\[ s = \frac{3 \times 6.02 \times 1.38}{27} \approx \frac{24.91}{27} \approx 0.923 \, \text{J/(kg K)} \]
Converting to appropriate units:
\[ s \approx 923.06 \, \text{J/(kg K)} \]

### Step 8: Final Result
Thus, the specific heat of aluminum at room temperature is approximately:
\[ s \approx 925 \, \text{J/(kg K)} \]