The isentropic relation between the stagnation temperature ($T_{0}$) and the static temperature ($T$) is given by:
$$\frac{T_{0}}{T} = \left(1 + \frac{k-1}{2}M^2\right)$$
where $k$ is the specific heat ratio of the exhaust gases, and $M$ is the Mach number.
At the throat, the Mach number is 1, so we have:
$$\frac{T_{0}}{T_t} = \left(1 + \frac{k-1}{2}\right)$$
where $T_t$ is the static temperature at the throat.
We are also given the stagnation pressure ($P_0$) and the static pressure at the throat ($P_t$) of 4 MPa and, we can use the isentropic relation between pressure and temperature to find $T_t$:
$$\frac{P_0}{P_t} = \left(\frac{T_0}{T_t}\right)^{\frac{k}{k-1}}$$
Substituting the expression for $T_0/T_t$ from before, we get:
$$\frac{P_0}{P_t} = \left(1 + \frac{k-1}{2}\right)^{\frac{k}{k-1}}$$
Solving for $T_t$, we get:
$$T_t = \frac{P_t}{P_0}\left(1 + \frac{k-1}{2}\right)^{\frac{1}{1-k}}$$
Assuming the exhaust gases are ideal with $k = 1.4$ and $P_t = P_{exit}$ (since the flow is choked), we can calculate $T_t$:
$$T_t = \frac{101.325\text{ kPa}}{4000\text{ kPa}}\left(1 + \frac{0.4}{2}\right)^{\frac{1}{0.4}} \approx 712.71 \text{ K}$$
Now, we can use the isentropic relation between the stagnation temperature and the static temperature again to find the stagnation temperature $T_0$:
$$T_0 = \left(1 + \frac{k-1}{2}\right)T_t$$
$$T_0 = \left(1 + \frac{0.4}{2}\right)(712.71 \text{ K}) \approx 1068.77 \text{ K}$$
Therefore, the stagnation temperature at the combustion chamber is approximately 1069 K.
