= Solution
\b[Exchange constraint.] Exchange of two <identical particles> cannot change physical predictions. For the usual three-dimensional bosonic or fermionic state space, the <exchange operator> obeys $P_{12}^2=1$, so its allowed <eigenvalues> are $\pm1$. <bosons> use the symmetric overall <wavefunction>, $P_{12}\Psi=+\Psi$; <fermions> use the antisymmetric overall <wavefunction>, $P_{12}\Psi=-\Psi$. Exchange includes the spatial coordinates and all internal labels. The identity $P_{12}^2=1$ alone does not prove which particle species has which statistics; that assignment is the <Spin-statistics theorem>.
\b[Factored <fermion> states.] If $\Psi=\psi_{\rm space}\chi_{\rm spin}$ and both factors have definite exchange symmetry, their exchange signs must multiply to $-1$. Thus a symmetric spatial factor requires an antisymmetric spin factor, and an antisymmetric spatial factor requires a symmetric spin factor. For two spin-one-half particles, the former spin factor is the singlet $(|\uparrow\downarrow\rangle-|\downarrow\uparrow\rangle)/\sqrt2$; the latter factors are the three <triplet states>. A more general entangled spin-space state need not factor, but must still be antisymmetric overall.
\b[<hydrogen> spin isomers.] The two <protons> are identical spin-one-half <fermions>. <parahydrogen> has total <nuclear spin> $I_{\rm n}=0$, the antisymmetric singlet; <orthohydrogen> has $I_{\rm n}=1$, the symmetric triplet. Here $I_{\rm n}$ is a spin quantum number, distinct from the <moment of inertia> used below.
\b[Rotational restriction.] In the electronic and vibrational <ground state>, these two factors are symmetric under nuclear exchange. Exchanging the <protons> reverses the internuclear axis, $(\theta,\phi)\mapsto(\pi-\theta,\phi+\pi)$. The <spherical harmonic> rotational state contributes the exchange sign $(-1)^\ell$. Overall antisymmetry therefore requires
$$
(-1)^\ell(-1)=-1\quad\hbox{for <parahydrogen>},\qquad
(-1)^\ell(+1)=-1\quad\hbox{for <orthohydrogen>}.
$$
Hence \b[<parahydrogen> has even $\ell$; <orthohydrogen> has odd $\ell$]. The $\ell$ here labels <molecular rotation>, rather than the <angular momentum> obtained by also coupling the <nuclear spins>.
\b[Conversion <heat>.] The <ortho-para conversion heat> follows from the <linear rigid rotor> approximation, $E_\ell=\hbar^2\ell(\ell+1)/(2I)$ with $I=\mu r^2\simeq m_pr^2/2$. Thus conversion of one <molecule> from $\ell=1$ to $\ell=0$ releases
$$
\Delta E=\frac{\hbar^2}{I}=\frac{2\hbar^2}{m_pr^2}
\simeq2.43\times10^{-21}\ \mathrm J.
$$
Neglecting the small <electron> contribution to the molecular mass, a sample of mass $M$ contains $M/(2m_p)$ <molecules>. Only the initial <orthohydrogen> fraction releases this <energy>, giving
$$
Q=\frac34\frac{M}{2m_p}\frac{2\hbar^2}{m_pr^2}
=\frac{3M\hbar^2}{4m_p^2r^2}
=\boxed{5.44\times10^5\ \mathrm J\quad(M=1\,\mathrm{kg})}.
$$
The estimate uses $r=74$ pm, $m_p\simeq1.67262\times10^{-27}$ kg and $\hbar\simeq1.05457\times10^{-34}$ J s. Handbook-level rounding and the <rigid rotor> approximation justify about three significant figures, not a high-precision thermochemical prediction.<CODATA recommended values of the fundamental physical constants: 2022 by National Institute of Standards and Technology>[recommended-value entries: proton mass and reduced Planck constant]{ref}
\b[Storage consequence.] The internal conversion is a delayed <heat> source even in a well insulated container, causing warming or additional evaporation of the liquid. Insulation reduces external <heat transfer> but does not remove this source. Converting to predominantly <parahydrogen> during liquefaction, while removing the released <heat>, reduces the subsequent storage <heat> load. No evaporation fraction can be calculated without a <latent heat> and a model for how the <heat> is removed.
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