Is $E′(k,m)=E(0^n,m)$ semantically secure?

$(E,D)$ is a (one-time) semantically secure cipher where the message and ciphertext space is $\{0,1\}^n$. $E'(k,m)$ is defined as

$E′(k,m)=E(0^n,m)$

The question is whether $E'(k,m)$ is (one-time) semantically secure?

I thought that the random chosen key $k$ might be $0^n$, therefore the adversary cannot distinguish the messages $m0$ and $m1$ when encrypted with the key $0^n$. Therefore, the scheme is semantically secure. However, the answer states that

To break semantic security, an attacker would ask for the encryption of $0^n$ and $1^n$ and can easily distinguish $EXP(0)$ from $EXP(1)$ because it knows the secret key, namely $0^n$.

Can you elaborate on the answer more? I don't understand how the key is known ("because it knows the secret key") and how we can find that the messages $0^n$ and $1^n$ should be used to break the semantic security (Can we try the whole message space?)?

• If you hard-code the key it's part of the algorithm and by Kerckhoff's principle the attacker knows this then, so he can just try it.
– SEJPM
May 14 '16 at 13:33
• That makes the key issue clear, thanks! Does this mean that we don't have to use $0^n$ and $1^n$ to break the semantic security, we can actually use any two different messages to break the semantic security?
– sha1
May 14 '16 at 13:40
• Possible duplicate of Is the identity function a one-way function? May 14 '16 at 16:02
• Maybe we need a "master question" about this. Preferably one which does not involve the quite vague statement known as Kerckhoff's principle, which causes more confusion than it solves. May 14 '16 at 16:06

The algorithm $E'(m)=E'(k,m)=E(0^n,m)$ is defined with a hard-coded key, thus the key is part of the algorithm definition of $E'$.
The choice of the messages $0^n$ and $1^n$ is entirely arbitrary and any message would suit the purpose here with these two having the advantage of the shortest possible description (instead of something like $0^{n-1}||1$).