Question

What is the thermal energy $100c{m}^3$ of aluminum at ${100}^{0}c$ ?

Short answer

The thermal energy of aluminum at ${100}^{0}C$is$E_{\text{th}}=93.1\mathrm{kJ}$

Step-by-step solution

Step :1 Introduction 

In the microsystem, the potential energy between the bonds is zero, and the kinetic energy of the system is translational kinetic energy. As a result, a solid's thermal energy is $n$moles is given by equation (20.39)in the form

$E_{\mathrm{th}}=3nRT$

The state variables that describe the state of the system are used to describe a macroscopic system. One of the state variables is density, which is determined by the ratio of the other two state variables (mass and volume) and may be calculated using equation (18.1) in the form

$\rho =\frac{M}{V}$

Step  :2 Explanation 

Knowing the mass$M$and the molar mass $m$, we can get the number of moles by

$M=nm$

We substitute the number of moles $n$into equation (1) to get the thermal energy by

$E_{\mathrm{th}}=3\left(\frac{\rho V}{m}\right)RT$

Step :3  Substitution 

The density of aluminum is $\rho =2.7\mathrm{g}/{\mathrm{cm}}^3$, the molar mass is $m=27\mathrm{g}/\mathrm{mol}$. Now, we plug the values for and $T$into equation (3) to get $E_{\text{th}}$

$E_{\mathrm{th}}=\left(\frac{3\rho VRT}{m}\right)$

$=\left(\frac{3\left(2.7\mathrm{g}/{\mathrm{cm}}^3\right)\left(100{\mathrm{cm}}^3\right)(8.314\mathrm{J}/\mathrm{K}\cdot \mathrm{mol})\left({373}^{\circ }\mathrm{K}\right)}{27\mathrm{g}/\mathrm{mol}}\right)$

$\begin{array}{r}=93.1\times {10}^3\mathrm{J}\\ \end{array}$

$=93.1KJ$