= Solution
Take positive $\dot Q$ to mean net removal of thermal <energy> per unit mass. The <isobaric thermal stability of a heated ideal gas> depends on the density change accompanying a temperature disturbance. For an <ideal gas> of fixed composition, $p=\rho k_BT/\bar m$, so constant <pressure> gives $\rho T=\text{constant}$. Therefore the net cooling can be written
$$
\mathcal L(T)=CT^{\alpha-1}-H,
$$
where $C>0$ is constant for this pressure. At thermal equilibrium $T=T_0$, $CT_0^{\alpha-1}=H$.
The <first law of thermodynamics> at constant pressure includes the expansion work, so the relevant heat capacity is the <specific heat capacity at constant pressure>, $c_p$, rather than the constant-volume value. Linearizing its temperature equation gives
$$
c_p\frac{d\,\delta T}{dt}
=-\left(\frac{\partial\mathcal L}{\partial T}\right)_p\delta T
=-\frac{(\alpha-1)H}{T_0}\delta T.
$$
Hence
$$
\boxed{\delta T\propto e^{st},\qquad
s=\frac{(1-\alpha)H}{c_pT_0}.}
$$
For optically thin <Bremsstrahlung>, $\alpha=1/2$, so
$$
\boxed{s=\frac{H}{2c_pT_0}>0:\quad\text{the equilibrium is thermally unstable}.}
$$
A hotter parcel expands and becomes less dense. Its cooling per unit mass, proportional to $T^{-1/2}$ at fixed pressure, decreases while its <cosmic ray> heating stays fixed, making it hotter still. A colder, denser parcel cools faster and runs away in the opposite direction. Here $H$ has units $\mathrm{W\,kg^{-1}}$ and $c_pT_0$ units $\mathrm{J\,kg^{-1}}$, so $s$ is an inverse time. The conclusion assumes the disturbance remains close to pressure equilibrium and neglects <heat conduction> and changes of composition. The relevant $\alpha$ is the temperature exponent in the supplied cooling law.
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