Let $E_t:=e^{t\alpha^2\Delta}$ be the solution operator to the heat equation
$$ \frac{\partial \Theta}{\partial t} = \alpha^2 \Delta \Theta$$
subject to the initial and boundary contitions above. Then the solution to your equation is given by

$$ \Theta_t = \Theta_0 +\int_0^t E_{t-s} T \,\mathrm{d} s.$$

Here the term $T$ is your distributional right-hand side.

Regarding well-posedness: Notice that $\Theta_t$ is at least a distributional solution: For any test-function $\varphi$, we have
$$\partial_t \langle \Theta_t, \varphi\rangle = \langle T, \varphi\rangle + \int_0^t \langle T, E_{t-s} \Delta \varphi\rangle \mathrm{d} s$$
so that $\Theta_t$ solves the equation in the distributional sense.

Regarding the Sobolev regularity: For $T$ regular enough, you have
$$|\langle \Delta E_t T, \varphi\rangle| \leq \|\Delta E_t T\|_{H^{-s}}\|\varphi\|_{H^s} \leq C \|E_t T\|_{H^{2-s}}\|\varphi\|_{H^s}$$
The heat operator has the short-time asymptotics
$$\|E_t \|_{H^{r}, H^l} \leq C t^{(r-l)/2}.$$
for $l>r$. Here $l=2-s$ so that we obtain
$$|\langle \Delta E_t T, \varphi\rangle| \leq C^\prime t^{(r-2+s)/2}\|T\|_{H^r}\|\varphi\|_{H^{s}} = C^\prime t^{-1+r/2+s/2}\|T\|_{H^r}\|\varphi\|_{H^{s}}$$
In order for this to be integrable at zero, you want the exponent to be greater than $-1$. In your special case, we have $T \in H^r$ for every $r=-n/2-\varepsilon$ with $\varepsilon>0$ so that we can choose $s>n/2$ (here $n$ is the dimension of your domain).

What does this tell us? The functional
$$\langle \Theta_t, - \rangle$$
is bounded for any $s>n/2$, hence $\Theta_t \in H^{-s}$ for every such $s$.