Question

Two farmers are standing on opposite sides of a very large empty field that is \(510 . \mathrm{m}\) across. One farmer yells out some instructions, and \(1.50 \mathrm{~s}\) pass until the sound reaches the other farmer. What is the temperature of the air?

Short answer

Answer: The temperature of the air is approximately \(11.96^{\circ}\mathrm{C}\).

Step-by-step solution

Find the speed of sound

Use the formula for speed, which is \(v = \frac{d}{t}\), where \(v\) is the speed of sound, \(d\) is the distance, and \(t\) is the time. We are given the distance (\(510 \mathrm{m}\)) and the time (\(1.50 \mathrm{s}\)). Plugging these values into the formula, we get: $$v = \frac{510 \mathrm{m}}{1.50 \mathrm{s}}$$ $$v = 340 \mathrm{\frac{m}{s}}$$

Solve for the temperature

Now that we have the speed of sound, we can use the formula for the speed of sound in air to find the temperature. The formula is: $$v = 331.4\sqrt{1+\frac{T}{273}}$$ We know the speed of sound \(v = 340 \mathrm{\frac{m}{s}}\), and we want to find the temperature \(T\). First, we will isolate the temperature in the equation: $$\frac{v}{331.4} = \sqrt{1+\frac{T}{273}}$$ Now, square both sides to get rid of the square root: $$\left(\frac{v}{331.4}\right)^2 = 1+\frac{T}{273}$$ Substitute the value of \(v\): $$\left(\frac{340}{331.4}\right)^2 = 1+\frac{T}{273}$$ Now, solve for \(T\): $$T = 273\left(\left(\frac{340}{331.4}\right)^2 - 1\right)$$ $$T \approx 11.96 ^{\circ}\mathrm{C}$$ The temperature of the air is approximately \(11.96 ^{\circ}\mathrm{C}\).