Question

The temperature at the cloud tops of Saturn is approximately \(150 .\) K. The atmosphere of Saturn produces tremendous winds; wind speeds of \(600 . \mathrm{km} / \mathrm{h}\) have been inferred from spacecraft measurements. Can the wind chill factor on Saturn produce a temperature at (or below) absolute zero? How, or why not?

Short answer

Answer: Based on the approximation using a simple wind chill formula and the given conditions, the wind chill factor on Saturn cannot produce a temperature at or below absolute zero. The calculated wind chill temperature is approximately 66.67 K, which is greater than 0 K.

Step-by-step solution

Convert wind speeds to meters per second

To convert the wind speed from km/h to m/s, we can use the following conversion formula: 1 km/h = 1000 m / 3600 s = 5/18 m/s So, the wind speed of 600 km/h is equivalent to: (600 km/h) * (5/18 m/s per km/h) = (600 * 5/18) m/s

Calculate the wind chill temperature

We will use the basic wind chill formula to calculate the wind chill temperature. This formula is given by: \(T_{windchill} = T_{0} + k \cdot v\) Where: - \(T_{windchill}\) is the wind chill temperature - \(T_{0}\) is the initial temperature (in this case, 150 K), - \(v\) is the wind speed (in m/s), and - \(k\) is a constant (in K / m/s) Note that the wind chill formula may considerably vary depending on the circumstances, so our result is just an approximation. For this exercise, we assume that the value of \(k\) in the given circumstances is -0.4 K / m/s (keep in mind that the real wind chill formulas also consider other factors, which might not be provided in this exercise). Now we can plug the values into the formula: \(T_{windchill} = 150 K + (-0.4 K / m/s) \cdot (600 * 5/18) m/s\)

Determine if the wind chill temperature reaches absolute zero or below

Now, we need to see if the wind chill temperature calculated in the previous step reaches absolute zero (0 K) or below. To do this, simply solve the equation above: \(T_{windchill} = 150 K -0.4 K / m/s \cdot (600 * 5/18) m/s = 150 K - 0.4 \cdot (600 * 5/18) K\) Using a calculator or appropriate software, we can find the wind chill temperature: \(T_{windchill} = 150 K - 0.4 \cdot (600 * 5/18) K \approx 66.67 K\) Since 66.67 K is greater than 0 K, the wind chill factor on Saturn cannot produce a temperature at or below absolute zero in this approximation.