= Solution
Assume an ideal isentropic turbine and use the supplied table consistently. Its inlet <specific entropy> is $s_3=6.282\,\mathrm{kJ\,kg^{-1}\,K^{-1}}$. At the condenser pressure the outlet mixture has $s_4=s_f+\chi(s_g-s_f)$, where $\chi$ is the vapour mass fraction. Hence
$$
\boxed{\chi=\frac{6.282-0.297}{8.667-0.297}=0.71505.}
$$
The outlet is approximately $71.5\%$ vapour and $28.5\%$ liquid by mass. Its <specific enthalpy> is
$$
h_4=84+0.71505(2538-84)=1838.74\,\mathrm{kJ\,kg^{-1}}.
$$
The source's common unit label omits the kelvin denominator for entropy: $s$ must be interpreted in $\mathrm{kJ\,kg^{-1}\,K^{-1}}$, while $h$ is in $\mathrm{kJ\,kg^{-1}}$. A real irreversible turbine would increase entropy and require an efficiency correction; no such correction is supplied.
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