Atmospheric Temp
200 meters above the surface of the Earth, the local speed of sound is 342.8 m/s. If the ratio of specific heats is 1.3, what is the atmospheric air temperature (°C)? Assume the molecular weight of air is 29 kg/kmol, and note the universal gas constant is 8,314 J/(kmol·K).
Expand Hint
The local speed of sound in an ideal gas is:
$$$c=\sqrt{kRT}$$$
where
$$k$$
is the ratio of specific heats,
$$R$$
is the specific gas constant, and
$$T$$
is the absolute temperature.
Hint 2
To find
$$R$$
:
$$$R=\frac{\bar{R}}{(mol.wt_i)}$$$
where
$$\bar R$$
is the universal gas constant.
To find the local speed of sound in an ideal gas:
$$$c=\sqrt{kRT}$$$
where
$$k$$
is the ratio of specific heats,
$$R$$
is the specific gas constant, and
$$T$$
is the absolute temperature. To find
$$R$$
:
$$$R=\frac{\bar{R}}{(mol.wt_i)}$$$
where
$$\bar R$$
is the universal gas constant. Thus,
$$$R=\frac{8,314J(kmol)}{(kmol\cdot K)(29kg)}=286.69\frac{J}{kg\cdot K}=286.69\:\frac{m^2}{s^2\cdot K}$$$
Solving for absolute temperature:
$$$342.8\frac{m}{s}=\sqrt{(1.3)(286.69\frac{m^2}{s^2 \cdot K})T}$$$
$$$(342.8\frac{m}{s})^2=372.697\frac{m^2}{s^2 \cdot K}\cdot T$$$
$$$T=\frac{117,511.84K}{372.697}=315.3\:K$$$
Finally, converting the temperature to Celsius:
$$$T=315.3-273=42.3^{\circ}C$$$
42.3°C
Time Analysis
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