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Convert the following to equivalent temperatures on the Celsius and Kelvin scales: (a) the normal human body temperature, \(98.6^{\circ} \mathrm{F} ;\) (b) the air temperature on a cold day, \(-5.00^{\circ} \mathrm{F}\).

Short Answer

Expert verified
The normal human body temperature is \(37.0^{\circ} C\) or \(310.15K\). The air temperature on a cold day is \(-20.56^{\circ} C\) or \(252.59K\)

Step by step solution

01

Convert Normal Body Temperature to Celsius

First, substitute \(F = 98.6\) into the formula for Celsius: \(C = (98.6 - 32) \times \frac{5}{9} \). Calculate the result.
02

Convert Normal Body Temperature to Kelvin

Now, substitute \(F = 98.6\) into the formula for Kelvin: \(K = (98.6 - 32) \times \frac{5}{9} + 273.15\). Calculate the result.
03

Convert Air Temperature to Celsius

Next, substitute \(F = -5.00\) into the formula for Celsius: \(C = (-5.00 - 32) \times \frac{5}{9}\). Calculate the result.
04

Convert Air Temperature to Kelvin

Finally, substitute \(F = -5.00\) into the formula for Kelvin: \(K = (-5.00 - 32) \times \frac{5}{9} + 273.15\). Calculate the result.

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Key Concepts

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

Celsius to Kelvin Conversion
Understanding the conversion between Celsius and Kelvin is pivotal in the field of physics, especially when dealing with temperature-related experiments or calculations. The Celsius scale is commonly used for everyday temperature measurements, while the Kelvin scale is the standard unit of measurement for thermodynamic temperature in the International System of Units (SI).

To convert from Celsius to Kelvin, you simply add 273.15 to the Celsius temperature. This is because the size of one degree Celsius is equal to the size of one Kelvin, but the two scales have different starting points: 0°C corresponds to the freezing point of water, whereas 0K (or absolute zero) is the point at which particles have the least thermal motion possible.

For example, if you have a temperature of 25°C and want to convert it to Kelvin, the calculation is straightforward:
  • Add 273.15 to the Celsius temperature: 25 + 273.15 = 298.15 K
This simple addition transitions your measurement from the Celsius scale to the Kelvin scale, aligning with the absolute temperature scale used in scientific contexts.
Fahrenheit to Celsius Conversion
Fahrenheit to Celsius conversion is necessary in many scientific contexts as the Celsius scale is more universally accepted in the scientific community, despite Fahrenheit being prevalent in some countries for everyday use. To convert a temperature from Fahrenheit to Celsius, we follow a two-step process that requires subtraction followed by multiplication.

The formula is:
  • Celsius temperature (°C) = (Fahrenheit temperature (°F) - 32) × 5/9
The subtraction of 32 degrees adjusts for the difference in the freezing point of water between the two scales (32°F in Fahrenheit is equivalent to 0°C in Celsius), and the multiplication by 5/9 converts the Fahrenheit degree increments to Celsius degree increments.

For instance, to convert 98.6°F (a typical human body temperature) to Celsius:
  • Subtract 32 from the Fahrenheit temperature: 98.6 - 32 = 66.6
  • Multiply the result by 5/9: 66.6 × 5/9 ≈ 37°C
Through these steps, you can accurately translate temperatures from the Fahrenheit scale to the Celsius scale.
Thermodynamic Temperature Scales
The significance of thermodynamic temperature scales in physics cannot be overstated as they are crucial for scientific accuracy and consistency. Thermodynamic temperature is an absolute measure of temperature, independent of the properties of any particular materials.

The Kelvin scale is the SI unit for thermodynamic temperature, and it is an absolute scale based on absolute zero—the theoretical temperature at which particles within a system have the minimum thermal motion and cannot be any colder. Another common thermodynamic scale is the Celsius scale, directly related to Kelvin but offset to set the freezing point of water at 0°C.

These scales are foundational for understanding the behavior of systems in thermodynamics. They allow scientists and engineers to calculate the absolute energy within a system and apply universal laws such as the ideal gas law and principles of heat transfer. Whether it is in experimental particle physics, engineering systems, or meteorology, being fluent with thermodynamic temperature scales like Celsius and Kelvin ensures precise communication and calculation of thermal energy states.

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Most popular questions from this chapter

Helium gas is sold in steel tanks. If the helium is used to inflate a balloon, could the balloon lift the spherical tank the helium came in? Justify your answer. Steel will rupture if subjected to tensile stress greater than its yield strength of \(5 \times 10^{8} \mathrm{N} / \mathrm{m}^{2} .\) Suggestion: You may consider a steel shell of radius \(r\) and thickness \(t\) containing helium at high pressure and on the verge of breaking apart into two hemispheres.

Liquid nitrogen has a boiling point of \(-195.81^{\circ} \mathrm{C}\) at atmospheric pressure. Express this temperature (a) in degrees Fahrenheit and (b) in kelvins.

One mole of oxygen gas is at a pressure of 6.00 atm and a temperature of \(27.0^{\circ} \mathrm{C} .\) (a) If the gas is heated at constant volume until the pressure triples, what is the final temperature? (b) If the gas is heated until both the pressure and volume are doubled, what is the final temperature?

The pressure gauge on a tank registers the gauge pressure, which is the difference between the interior and exterior pressure. When the tank is full of oxygen \(\left(\mathrm{O}_{2}\right),\) it contains \(12.0 \mathrm{kg}\) of the gas at a gauge pressure of \(40.0 \mathrm{atm} .\) Determine the mass of oxygen that has been withdrawn from the tank when the pressure reading is 25.0 atm. Assume that the temperature of the tank remains constant.

On a Strange temperature scale, the freezing point of water is \(-15.0^{\circ} \mathrm{S}\) and the boiling point is \(+60.0^{\circ} \mathrm{S} .\) Develop a linear conversion equation between this temperature scale and the Celsius scale.

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