Bitcoin mining: how is a block header of 80 bytes processed in SHA-256? Isn't it too big?
The SHA-256 compression function takes a 256-bit chaining value $h$ and a 512-bit message block $m$ and returns a 256-bit chaining value $f(h, m)$.

The SHA-256 hash function takes a padded message $m$ broken into an integral number of 512-bit message blocks $m = m_1 \mathbin\| m_2 \mathbin\| \cdots \mathbin\| m_\ell$, up to $\ell \leq 2^{55} + 1$ blocks long (the unpadded message length in bits, encoded into the last block, is 64 bits, allowing up to $\lceil(2^{64} - 1)/512\rceil = 2^{55}$ message blocks plus one more for the padded length), and iterates the compression function using a standard initialization vector $\mathit{iv}$ as the initial chaining value: $$H(m) = f(\cdots f(f(\mathit{iv}, m_1), m_2)\cdots, m_\ell).$$

If you are considering messages of the form $m_1 \mathbin\| m_2^i$ where you vary $m_2^i$ for the $i^{\mathit{th}}$ iteration, then of course you can precompute $h_1 = f(\mathit{iv}, m_1)$ and then use only a single evaluation of the SHA-256 compression function to compute $H(m_1 \mathbin\| m_2^i)$ by $f(h_1, m_2^i)$.