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\text { The surface temperature of the Sun is about } 5750 \mathrm{~K} \text { . What is this temperature on the Fahrenheit scale? }

Short Answer

Expert verified
The approximate surface temperature of the Sun in Fahrenheit is 9903 \textdegree F.

Step by step solution

01

Identify the Conversion Formula

To convert a temperature from Kelvin (K) to Fahrenheit (\textdegree F), use the formula: \( F = \frac{9}{5} (K - 273) + 32 \) where F is the Fahrenheit temperature and K is the Kelvin temperature.
02

Plug in the Given Kelvin Temperature

Insert the known temperature in Kelvin into the formula. Given that the surface temperature of the Sun is about 5750 K, the formula becomes: \( F = \frac{9}{5} (5750 - 273) + 32 \)
03

Perform the Calculation

First, subtract 273 from 5750 to convert Kelvin to Celsius: \( 5750 - 273 = 5477 \). Next, multiply this result by \( \frac{9}{5} \) and then add 32 to convert Celsius to Fahrenheit: \( F = \frac{9}{5} \times 5477 + 32 \approx 9871 + 32 = 9903 \) The approximate Fahrenheit temperature is 9903.

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Key Concepts

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

Kelvin to Fahrenheit Conversion
Understanding how to convert temperatures from the Kelvin scale to the Fahrenheit scale is crucial for several scientific and engineering applications. The formula for this conversion is a simple linear equation: \( F = \frac{9}{5} (K - 273) + 32 \), where \( F \) represents the Fahrenheit temperature and \( K \) represents the Kelvin temperature. It's important to remember that 273 is the difference between the Kelvin and Celsius scales, and the factors \( \frac{9}{5} \) and 32 are used to adjust the Celsius scale to the Fahrenheit scale.

Let's break down the Kelvin to Fahrenheit conversion into easy steps using our example with the Sun's surface temperature of 5750 Kelvin. When you input our given Kelvin value into the formula, the calculation starts by converting Kelvin to Celsius through subtracting 273. Next, we convert Celsius to Fahrenheit by multiplying the Celsius temperature by \( \frac{9}{5} \) and then adding 32. This results in an astonishing 9903 degrees Fahrenheit, reflecting the incredibly high temperatures the Sun's surface emits.

In a practical sense, this conversion is essential for understanding various scientific phenomena, from studying the vast temperatures in space to practical applications in industries that depend on precise temperature measurements.
Thermodynamic Temperature Scale
The thermodynamic temperature scale, known as the Kelvin scale, is a fundamental concept in physics and engineering. Named after the physicist Lord Kelvin, it is an absolute temperature scale, meaning it starts at absolute zero. Absolute zero, or 0 Kelvin, is theoretically the lowest possible temperature, where all thermal motion ceases in a perfect crystal lattice.

The Kelvin scale is a direct measure of thermal energy in a substance. Unlike the Celsius and Fahrenheit scales, it does not use degrees; temperatures are simply stated in kelvins. This scale plays a vital role in areas like thermodynamics, where understanding the absolute measure of temperature is necessary for calculating thermodynamic properties and predicting the behavior of matter under various thermal conditions. For instance, in our exercise, the high Kelvin temperature of the Sun is indicative of the immense thermal energy present on its surface, confirming it as a powerful source of heat and light for our solar system.
Physics Temperature Calculations
Temperature calculations in physics are crucial for a wide array of scientific inquiries and technological developments. In physics, temperature is a measure of the average kinetic energy of the particles in a substance. Temperature can influence the physical state, reaction rate, and equilibrium of chemical reactions, as well as the emission of radiation in astrophysics.

Working with temperature in physics often requires a firm grasp of various conversion formulas to transition between temperature scales, such as Kelvin, Celsius, or Fahrenheit. Knowing how to perform these conversions is essential for interpreting experimental data, operating in different units, and communicating results in universally understood terms. For example, while astronomers may use the Kelvin scale to measure stellar temperatures, meteorologists might convert those readings to Fahrenheit for public weather reports. The Sun's high surface temperature, as calculated in the given exercise, not only holds significance for astrophysical studies but also has implications for understanding the Earth's climate and the balances within our solar system.

By mastering these conversions, students and professionals can ensure accuracy in their work, whether it's in a laboratory setting, industrial applications, or theoretical physics. As temperatures can vary significantly in different environments and scales, being comfortable with these calculations is a key skill in the field of physics.

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Most popular questions from this chapter

(a) If the rocket sled shown in Figure 4.31 starts with only one rocket burning, what is the magnitude of its acceleration? Assume that the mass of the system is \(2100 \mathrm{kg}\), the thrust \(\mathrm{T}\) is \(2.4 \times 10^{4} \mathrm{N},\) and the force of friction opposing the motion is known to be 650 N. (b) Why is the acceleration not onefourth of what it is with all rockets burning?

A 35.0-kg dolphin decelerates from 12.0 to 7.50 m/s in 2.30 s to join another dolphin in play. What average force was exerted to slow him if he was moving horizontally? (The gravitational force is balanced by the buoyant force of the water.)

Because humidity depends only on water's vapor pressure and temperature, are the saturation vapor densities listed in Table valid in an atmosphere of helium at a pressure of \(1.01 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}\), rather than air? Are those values affected by altitude on Earth?

Commercial airplanes are sometimes pushed out of the passenger loading area by a tractor. (a) An 1800 -kg tractor exerts a force of \(1.75 \times 10^{4} \mathrm{N}\) backward on the pavement, and the system experiences forces resisting motion that total \(2400 \mathrm{N} .\) If the acceleration is \(0.150 \mathrm{m} / \mathrm{s}^{2},\) what is the mass of the airplane? (b) Calculate the force exerted by the tractor on the airplane, assuming 2200 N of the friction is experienced by the airplane. (c) Draw two sketches showing the systems of interest used to solve each part, including the free-body diagrams for each.

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