Answer
$320.0$ K
$320.1$ K
$325.0$ K
$817.5$ K
Work Step by Step
The air which strikes the probe will be brought to a complete stop, and thus it will undergo a stagnation process. The thermometer will sense the temperature of this stagnated air, which is the stagnation temperature, $T_0$. It is determined from $T_0=T+\frac{V^2}{2 c_p}$. The results for each case are calculated below:
(a)
$$
T_0=320 \mathrm{~K}+\frac{(1 \mathrm{~m} / \mathrm{s})^2}{2 \times 1.005 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}}\left(\frac{1 \mathrm{~kJ} / \mathrm{kg}}{1000 \mathrm{~m}^2 / \mathrm{s}^2}\right)=\mathbf{3 2 0 . 0 K}
$$
(b)
$$
T_0=320 \mathrm{~K}+\frac{(10 \mathrm{~m} / \mathrm{s})^2}{2 \times 1.005 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}}\left(\frac{1 \mathrm{~kJ} / \mathrm{kg}}{1000 \mathrm{~m}^2 / \mathrm{s}^2}\right)=\mathbf{3 2 0 . 1 K}
$$
(c)
$$
T_0=320 \mathrm{~K}+\frac{(100 \mathrm{~m} / \mathrm{s})^2}{2 \times 1.005 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}}\left(\frac{1 \mathrm{~kJ} / \mathrm{kg}}{1000 \mathrm{~m}^2 / \mathrm{s}^2}\right)=\mathbf{3 2 5 . 0 K}
$$
(d)
$$
T_0=320 \mathrm{~K}+\frac{(1000 \mathrm{~m} / \mathrm{s})^2}{2 \times 1.005 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}}\left(\frac{1 \mathrm{~kJ} / \mathrm{kg}}{1000 \mathrm{~m}^2 / \mathrm{s}^2}\right)=\mathbf{8 1 7 . 5 K}
$$
