Answer
$c^{}= 308.0$ m/s
$V= 338.8$ m/s
$T_{0}= 293$ K
Work Step by Step
The speed of sound in air at the specified conditions is $$
c=\sqrt{k R T}=\sqrt{(1.4)(0.287 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K})(236.15 \mathrm{~K})\left(\frac{1000 \mathrm{~m}^2 / \mathrm{s}^2}{1 \mathrm{~kJ} / \mathrm{kg}}\right)}=308.0 \mathrm{~m} / \mathrm{s}
$$ Thus, $$
V=\mathrm{Ma} \times c=(1.1)(308.0 \mathrm{~m} / \mathrm{s})=338.8 \mathrm{~m} / \mathrm{s}
$$ Then, $$
T_0=T+\frac{V^2}{2 c_p}=236.15+\frac{(338.8 \mathrm{~m} / \mathrm{s})^2}{2 \times 1.005 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}}\left(\frac{1 \mathrm{~kJ} / \mathrm{kg}}{1000 \mathrm{~m}^2 / \mathrm{s}^2}\right)=293 \mathrm{~K}
$$
