Answer
See explanation.
Work Step by Step
(a) As the wind move fast, there is no heat exchange, which is expected to be in adiabatic process. According to the First Law of Thermodynamics, $ΔU = Q - W$ and when $Q = 0, ΔU = -W$. This means the work is done on the wind, hence the temperature will increase besides the internal energy.
(b) For adiabatic process,
$𝑇_1 𝑉_1^{𝛾−1}= 𝑇_2 𝑉_2^{𝛾−1}$
$𝑇_1^\gamma 𝑉_1^ {1-\gamma}= 𝑇_2^\gamma 𝑉_2^{1- 𝛾}$
Now we want to find the final temprerature, $T_2$
$T_2 = T_1 \frac{p_1}{p_2}^{(\gamma - 1)/ \gamma}$
$T_2 = (258 15 K) (\frac{8.12 \times 10^4 Pa}{5.60 \times 10^4 Pa})^{(1.4 - 1)/1.4} $
$T_2 = (258 15 K) (\frac{8.12 \times 10^4 Pa}{5.60 \times 10^4 Pa})^{(2/7)} $
$T_2 = 287.1 K = 13.95 ^o C$
The increase of temperature is
$\Delta T = T_2 - T$
$\Delta T =13.95 ^o C - 2.0 ^o C$
$\Delta T = 11.95 ^oC$
