How secure is XOR-CBC? How secure would this be? To be clear, I mean replacing the block cipher encryption step in the Wikipedia diagram explaining CBC with plain old XOR?

EDIT:

Replace block cipher encryption part of diagram with XOR ($\oplus$)

CBC diagram

Or:
($C_0 = IV$)

$$C_i \equiv \left(\bigoplus_{C_{i-1}}^{P_i}\right) \oplus {K}$$
Completely insecure. 

If you just replace the encryption with XOR with the key, then the first ciphertext block is just $c_1=m_1 \oplus k$. The second block would be $c_2=m_2 \oplus c_1 \oplus k = m_2 \oplus m_1$.

The second block is actually not encrypted at all. And that goes on, the first, third, fifth and so on block is just an XOR with the key, the second, fourth, sixth and so on doesn't depend on the key at all.