= Solution
Let $\Delta=\hbar\Omega$ and choose the nondegenerate <energy> levels as 0 and $\Delta$. For N <independent> identical units, $Z=(1+e^{-\beta\Delta})^N$; set N=1 for a single <two-level system>. Differentiating its logarithm gives
$$
\boxed{U=\frac{N\Delta}{1+e^{\Delta/(k_BT)}},\qquad
C_V=Nk_B\frac{x^2e^x}{(1+e^x)^2},\quad x=\frac\Delta{k_BT}.}
$$
At low <temperature>, $U\sim N\Delta e^{-x}$ and $C_V\sim Nk_Bx^2e^{-x}$: the upper level is frozen out. At high <temperature>, the two levels approach equal populations, giving $U\to N\Delta/2$ while $C_V\sim Nk_Bx^2/4\to0$. The <energy> therefore rises monotonically to a plateau, while the <heat capacity> has a Schottky peak:
$$
\boxed{k_BT_{\rm peak}\simeq0.417\Delta,\qquad C_{V,\rm peak}\simeq0.439Nk_B.}
$$
Unequal degeneracies would change the populations and peak; the usual equal-degeneracy model is stated here. Shifting both <energy> levels changes the <energy> baseline but not <heat capacity>.
\Image[/past-exam-of-the-thermal-and-statistical-physics-course-of-the-university-of-cambridge/2013/paper-1-tsp2013-two-level.png]
{title=Original equal-degeneracy two-level internal-energy and heat-capacity curves, with the high-temperature plateau and Schottky peak marked}
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