Answer
$ c_{1} = 342.9 m/s$
$ Ma_{1} = 0.292$
$T2 = 111C^o = 384K$
$ c_{2} = 399m/s$
$Ma2 = 0.501$
Work Step by Step
The gas const $R = 0.2968 KJ/(KG.K)$
The specific heat at constant pressure at constant temperature, $c_{p} = 1.040 KJ/(KG.K) $
Specific heat ratio , $K =1.4$
we have $c_{1} = \sqrt (K1RT1)$
Substituting the values
$ c_{1} = \sqrt(1.4 KJ/(KG.K) x 0.2968 KJ/(KG.K) x 283K x ((1000m2/s2)/(1KJ/Kg))$
on solving we get
$ c_{1} = 342.9 m/s$
We have $ Ma_{1} = v_{1}/c_{1} = (100 m/s)/(342.9 m/s) = 0.292$
We have the equation
$Q_{in} = c_{p}(T_{2}-T_{1}) + (v_{2}^2-v_{1}^2)/2$
given ,$ v_{2}=200m/s^2$, $v_{1}= 100m/s^2$, $Q_{in} = 120KJ/Kg$, T1= $10C^o$
substituriong the values we get
$120 Kj/kg = 1.040Kj/Kg C^o(T2 - 10C^o) + (((200 - 100)m/s2)/2 x ((1KJ/Kg)/(1000m2/s2)))$
$T2 = 111C^o = 384K$
we have $c_{2} = \sqrt(K2RT2)$
substituting the vaues
$c_{2} = \sqrt(1.4 KJ/(KG.K) x 0.2968 KJ/(KG.K) x 384K x ((1000m2/s2)/(1KJ/Kg))$
we get $ c_{2} = 399m/s$
similarly $Ma2 = v_{2}/c_{2}= (200 m/s)/(3399 m/s) = 0.501$
