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The temperature drops from \(60^{\circ} \mathrm{F}\) during the day to \(35^{\circ} \mathrm{F}\) during the night. (a) The corresponding temperature drop on the Celsius scale is (1) greater than, (2) the same as, or (3) less than. Explain. (b) Compute the temperature drop on the Celsius scale.

Short Answer

Expert verified
(a) Less than. (b) 13.89°C.

Step by step solution

01

Understand the Temperature Conversion Formula

To convert a temperature from Fahrenheit to Celsius, the formula is \( C = \frac{5}{9}(F - 32) \). We will use this formula to convert the given temperatures from Fahrenheit to Celsius.
02

Convert Day Temperature to Celsius

Convert the daytime temperature of \(60^{\circ} \mathrm{F}\) to Celsius using the formula. Substitute \( F = 60 \) into the formula: \[ C = \frac{5}{9} (60 - 32) = \frac{5}{9} \times 28 = \frac{140}{9} \approx 15.56^{\circ} \mathrm{C} \].
03

Convert Night Temperature to Celsius

Convert the nighttime temperature of \(35^{\circ} \mathrm{F}\) to Celsius. Substitute \( F = 35 \) into the formula: \[ C = \frac{5}{9} (35 - 32) = \frac{5}{9} \times 3 = \frac{15}{9} \approx 1.67^{\circ} \mathrm{C} \].
04

Calculate Temperature Drop in Celsius

Determine the difference in Celsius between the day and night temperatures. Subtract the night temperature from the day temperature: \[ 15.56^{\circ} \mathrm{C} - 1.67^{\circ} \mathrm{C} = 13.89^{\circ} \mathrm{C} \].
05

Compare the Temperature Drops

The drop in temperature in Fahrenheit is \(60 - 35 = 25^{\circ} \mathrm{F}\). In Celsius, the drop is \(13.89^{\circ} \mathrm{C}\). Since \(25^{\circ} \mathrm{F}\) is greater than \(13.89^{\circ} \mathrm{C}\), the drop in Celsius is less.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Fahrenheit to Celsius
When converting temperatures from Fahrenheit to Celsius, we rely on a specific conversion formula: \[ C = \frac{5}{9}(F - 32) \]. This formula helps us translate values between these two different temperature scales. Here's how it works:
  • The Fahrenheit temperature (\(F\)) is first adjusted by subtracting 32.
  • Then, the result is multiplied by \(\frac{5}{9}\) to convert it to Celsius (\(C\)).
This calculation adjusts for the fact that the zero points are different on each scale (0 degrees Celsius is freezing point of water, whereas 32 degrees Fahrenheit is). The ratio of \(\frac{5}{9}\) accounts for the different sizes of each degree on the respective scales.
Temperature Drop
A temperature drop is a simple subtraction that tells you how much the temperature has decreased over a certain period. In this exercise, the temperature decreases from \(60^{\circ} \mathrm{F}\) during the day to \(35^{\circ} \mathrm{F}\) at night. We calculate the drop in Fahrenheit first by subtraction: \[ 60 - 35 = 25^{\circ} \mathrm{F} \].
On the Celsius scale, which we calculated using the conversion formula, the equivalent drop is \[ 15.56^{\circ} \mathrm{C} - 1.67^{\circ} \mathrm{C} = 13.89^{\circ} \mathrm{C} \]. Notice that even though both calculate the same physical change in temperature, the numerical drop is different due to how each scale is structured.
Celsius Scale
The Celsius scale is often used globally, especially in scientific contexts. It's straightforward because it sets the freezing point of water at 0 degrees, and boiling at 100 degrees under normal atmospheric pressure. This decimal-based system is often easier to work with for scientific calculations compared to Fahrenheit.
In this exercise, using the Celsius scale means converting our original Fahrenheit measurements to see how the actual temperature drop compares numerically between the scales. This comparison is key:
  • 25 degrees on the Fahrenheit scale translates to a smaller numerical drop on the Celsius scale of approximately 13.89 degrees.
The difference illustrates how conversion affects temperature representation.
Scientific Calculations
Scientific calculations often require precision and consistent units, which is why temperature conversions are crucial. These conversions ensure that scientific experiments remain accurate and comparable worldwide. For temperature, converting from Fahrenheit to Celsius ensures compatibility with most scientific research formats.
The formula we used in this exercise (\[ C = \frac{5}{9}(F - 32) \]) is a prime example of how to maintain consistency when conducting analyses or reporting outcomes. In science, such conversions facilitate better understanding and prevent errors in interpretation. Additionally, knowledge of both scales might be essential in fields like meteorology or chemistry, where global communication is frequent.

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Most popular questions from this chapter

If the temperature of an ideal gas increases from \(300 \mathrm{~K}\) to \(600 \mathrm{~K},\) what happens to the rms speed of the gas molecules?

What is the average kinetic energy per molecule in a monatomic gas at (a) \(10^{\circ} \mathrm{C}\) and (b) \(90^{\circ} \mathrm{C} ?\)

The escape speed from the Earth is about \(11000 \mathrm{~m} / \mathrm{s}\) (Section 7.5). Assume that for a given type of gas to eventually escape the Earth's atmosphere, its average molecular speed must be about \(10 \%\) of the escape speed. (a) Which gas would be more likely to escape the Earth: (1) oxygen, (2) nitrogen, or (3) helium? (b) Assuming a temperature of \(-40^{\circ} \mathrm{F}\) in the upper atmosphere, determine the rms speed of a molecule of oxygen. Is it enough to escape the Earth? (Data: The mass of an oxygen molecule is \(5.34 \times 10^{-26} \mathrm{~kg}\), that of a nitrogen molecule is \(4.68 \times 10^{-26} \mathrm{~kg},\) and that of a helium molecule is \(6.68 \times 10^{-27} \mathrm{~kg}\).

During open-heart surgery it is common to cool the patient's body down to slow body processes and gain an extra margin of safety. A drop of \(8.5^{\circ} \mathrm{C}\) is typical in these types of operations. If a patient's normal body temperature is \(98.2^{\circ} \mathrm{F}\), what is her final temperature in both Celsius and Fahrenheit?

A steel-belted radial automobile tire is inflated to a gauge pressure of \(30.0 \mathrm{lb} / \mathrm{in}^{2}\) when the temperature is \(61^{\circ} \mathrm{F}\). Later in the day, the temperature rises to \(100^{\circ} \mathrm{F}\) Assuming the volume of the tire remains constant, what is the tire's pressure at the elevated temperature? [Hint: Remember that the ideal gas law uses absolute pressure.]

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