Is there is limit to what $2^{\aleph_0}$ can be in a countable transitive model?
How large can be the value of the continuum in a countable transitive (standard) model of ZFC? For instance, if we wanted to consider the case where $2^{\aleph_0} = \aleph_{\omega_1}$, where would the $\aleph_1$ smaller cardinals come from, after all, the model is only countable?
Assuming there exists any transitive model of ZFC, there exists a countable submodel by Lowenheim-Skolem. This countable submodel may not be transitive, but it is well-founded, so by Mostowski collapse it is isomorphic to a countable transitive model $M$.
Using forcing, you can then construct generic extensions of $M$ (which are also countable transitive models) in which $2^{\aleph_0}$ can be "arbitrarily large". More precisely, let us suppose for simplicity that $M$ satisfies GCH (this can be arranged by a class forcing, or by replacing $M$ with $L^M$). Then if $\kappa\in M$ is any ordinal such that $$M\vDash\text{"$\kappa$ is a cardinal of uncountable cofinality"},$$ there is a generic extension $M[G]$ of $M$ which has the same cardinals as $M$ and which satisfies $M[G]\vDash 2^{\aleph_0}=\kappa$. In particular, if (say) $M\vDash \kappa=\aleph_{\omega_1}$, then $M[G]$ is a countable transitive model in which $2^{\aleph_0}=\aleph_{\omega_1}$.
Note, however, that this does not mean that there actually are uncountably many cardinals in $M[G]$ below $2^{\aleph_0}$, any more than the fact that $M[G]\vDash 2^{\aleph_0}>\aleph_0$ means $M[G]$ actually contains uncountably many real numbers. What it means is that $M[G]$ thinks there are uncountably many cardinals less than what it thinks is $2^{\aleph_0}$. From the perspective of the outside universe, all of these "cardinals" below $2^{\aleph_0}$ are actually just countable ordinals (as is the cardinal $M$ thinks is $2^{\aleph_0}$), and there are only countably many of them. However, $M[G]$ doesn't know this, because no bijection from these sets to $\omega$ is an element of $M[G]$.
If you are still confused about this, you may find this Wikipedia page or the answers to this question helpful.