 Solution

At fixed chain length, each harmonic mode contributes a Bose–Einstein thermal occupation and a temperature-independent zero-point energy. The latter contributes no heat capacity. With $x=\hslash \omega /(k_{B}T)$,At low temperature only the approximately linear part of the acoustic dispersion is excited, so $g(\omega )\simeq 1/(\pi v_s)$. Using ${\int }_0^{\infty }{x}^2{e}^x/(e^x-1{)}^{2}dx={\pi }^2/3$ givesIt is linear in temperature, rather than the three-dimensional Debye $T^3$ law. At high temperature every scalar oscillator tends to $k_B$, henceThe cutoff temperature is ${\Theta }_m=\hslash {\omega }_m/k_B=305.5$ K. The curve rises smoothly from zero toward this plateau; ${\Theta }_m$ sets a crossover scale, not a sharp transition. In normalized variables its initial slope is $C_V/(N{k}_B)\simeq (2\pi /3)T/{\Theta }_m$. The low-temperature continuum limit also requires temperature above the finite-size level spacing; uniform center-of-mass translation is treated separately. Below the lowest nonzero mode energy a finite chain does not retain a continuum linear law down to arbitrary temperature.