Answer:
320.004975 K, 320.047 K , 324.975 K ,817.5124
Explanation:
Let air is an ideal gas
The temperature at [tex]c_p[/tex] is given by [tex]T_0={T+\frac{v^2}{2c_P}}[/tex]
(a) for speed of 1 m/sec [tex]T_0={T+\frac{v^2}{2c_P}}={320+\frac{1^2}{2\times 1005}}=320.0049[/tex]
(b) for speed of 10 m/sec[tex]T_0={T+\frac{v^2}{2c_P}}={320+\frac{10^2}{2\times 1005}}=320.0497 K[/tex]
(C) For speed of 100 m/sec [tex]T_0={T+\frac{v^2}{2c_P}}={320+\frac{100^2}{2\times 1005}}=324.975 K[/tex]
(d) For speed of 1000 m/sec [tex]T_0={T+\frac{v^2}{2c_P}}={320+\frac{1000^2}{2\times 1005}}=817.5124 K[/tex]
