Final answer:
The temperature of the air in the open tube is approximately 273 K. The correct answer is option C. 273 K.
Step-by-step explanation:
In this scenario, we can apply the formula for the frequency of a closed-open tube:
ƒ = (n * v) / (4 * L)
where ƒ is the frequency, n is the harmonic number (1 for the fundamental frequency), v is the speed of sound, and L is the length of the tube.
Given that the lowest resonance frequency is 662.6 Hz and the length of the tube is 25 cm (or 0.25 m), we can rearrange the formula to solve for v (speed of sound).
v = (4 * L * ƒ) / n
Substituting the given values, we find:
v = (4 * 0.25 * 662.6) / 1 = 662.6 m/s
Next, we can use the formula for the speed of sound to find the temperature of the air:
v = √(γ * R * T)
Where γ is the ratio of specific heats (1.4 for air), R is the gas constant (8.314 J/(mol·K)), and T is the temperature in Kelvin.
Rearranging the formula to solve for T, we get:
T = (v^2) / (γ * R) = (662.6^2) / (1.4 * 8.314) = 273 K
Therefore, the temperature of the air is approximately 273 K.
The correct answer is option C. 273 K.