To be more precise, suppose that $M$ is a model of ZF, for simplicity (or tactility) a set in some larger model $V$ of ZFC+Con(ZF), and suppose that $M\vDash``\alpha$ is an ordinal$"$. Must there be a model $N$ of ZFC (appearing in $V$) such that $\alpha\in N$ and an isomorphism of the partial structures given by the transitive closures of $\alpha$ in $M$ and $N$?


If yes, then can we also assume that $M$ and $N$ agree on cardinality for ordinals ${\leq}\alpha$?



Does every ordinal appearing in a model of ZF appear in a model of ZFC?
As Noah mentioned in the comments, the answer to the first question is "yes".


The answer to the second question is consistently "no", assuming a little consistency: if there is a transitive model of ZF in $V$, then there will be transitive models $M$ of ZF in $V$ and ordinals $\alpha$ of $M$ for which the extra requirement fails. For let $P$ be a countable transitive model of ZF of minimal height. Let $Q=L^P$, so $Q$ models ZFC and has the same ordinals as does $P$ (and $Q$ is also countable). Let $\lambda=\aleph\_\omega^Q$. So $Q\models$ "$\lambda$ is singular". Let $G$ be $(Q,\mathrm{Coll}(\omega,{<\lambda}))$-generic. Let $\mathbb{R}^\*$ be the resulting symmetric set of reals, i.e. $\mathbb{R}^\*=\bigcup\_{\alpha<\lambda}\mathbb{R}\cap Q[G\upharpoonright\alpha]$. Let $M=(L(\mathbb{R}^\*))^{Q[G]}$. Then $M\models$ ZF and $\mathbb{R}^\*=\mathbb{R}\cap M$ and $M\models$ "$\lambda=\omega\_1$", so $M\models$ "$\omega\_1$ is singular", since $Q=L^M\subseteq M$.


Now suppose there is a model $N$ of ZFC such that the ordinals of $N$ which are ${\leq\lambda}$ are (externally) isomorphic to those of $M$ which are ${\leq\lambda}$, and $M,N$ have the same cardinals $\leq\lambda$. Thus, $\lambda=\omega\_1^N$. But $L\_\lambda^N=L\_\lambda^M$, and the sequence $\left<\aleph\_n^{L^M}\right>\_{n<\omega}=\left<\aleph\_n^{L\_\lambda^M}\right>\_{n<\omega}$ is definable over $L\_\lambda^M$, so it is in $N$, so $N\models$ "$\lambda$ is singular", a contradiction.


(Without the assumption that $\lambda+1$ is wellfounded, it doesn't seem clear that we must get $L\_\lambda^N=L\_\lambda^M$, so the last paragraph seems to become problematic.)


