Does $H\vDash AC$

Question

The set $H_\kappa$ of sets hereditarily of cardinality less than $\kappa$ is defined as $H_\kappa=\{x||tc(x)|\lt\kappa\}$. What if we define the set $H=H_{Ord}$ of sets hereditarily of cardinality less than $Ord$; $H$ is the class of sets with some ordinal number as there cardinality. Equivalently, $H$ is the class of hereditarily well-ordered sets.

It is immediately obvious $V=H$ is equivalent to $AC$. My questions are as follows:

1. Is it true that $ZF\vdash H\vDash ZF$? Is it true that $ZF\vdash H\vDash ZFC$?

2. More generally, does $ZF\vdash H_\kappa\vDash AC$?

3. What is the consistency strength of the existence of some $j: H\prec H$? Is the existence of such an embedding first order expressible?

(I am obviously working in the context of $ZF$ without choice. I define $|tc(x)|\lt\kappa\leftrightarrow\exists\lambda\lt\kappa(|tc(x)|=\lambda)$)

Answer 1

No, yes, and not sure.

$H$ (and $H_\kappa$ in general) always satisfies choice, because any family of nonempty sets in $H$ has a well orderable transitive closure, from whence we can define a choice function, which is easily hereditarily well orderable as well.

However, $H\models\sf Power$ if and only if choice holds on $V$. Otherwise, let $\alpha$ be the least ordinal with a power set that cannot be well ordered, then $\mathcal P(\alpha)\subseteq H$ but it is not an element of $H$.