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(a) Calculate the one temperature at which Fahrenheit and Celsius thermometers agree with each other. (b) Calculate the one temperature at which Fahrenheit and Kelvin thermometers agree with each other.

Short Answer

Expert verified
Fahrenheit and Celsius agree at -40, and Fahrenheit and Kelvin agree at 574.59.

Step by step solution

01

Understand the Problem

We need to find two temperatures: one where the reading is the same in both the Fahrenheit and Celsius scales, and another where the reading is the same in both the Fahrenheit and Kelvin scales.
02

Formulate the First Equation

For part (a), we start with the formula that relates Celsius (\( C \) ) and Fahrenheit (\( F \) ) temperatures: \[ F = \frac{9}{5}C + 32 \] Since we need the temperatures to be equal, we set \( F = C \) and substitute in the formula: \[ C = \frac{9}{5}C + 32 \].
03

Solve the First Equation

Substituting into the equation \( C = \frac{9}{5}C + 32 \), we rearrange terms: \[ C - \frac{9}{5}C = 32 \] Simplify to get: \[ -\frac{4}{5}C = 32 \]. Multiply both sides by \(-\frac{5}{4}\) to solve for \( C \): \[ C = -40 \] Thus, \( F = C = -40 \).
04

Formulate the Second Equation

For part (b), Fahrenheit (\( F \)) and Kelvin (\( K \)) thermometers need to agree, which happens when \( F = K \). We use the formulas: 1. \( F = \frac{9}{5}(K-273.15) + 32 \)2. Since we want \( F = K \), we equate: \( \frac{9}{5}(K - 273.15) + 32 = K \).
05

Solve the Second Equation

Rearrange and solve \( \frac{9}{5}(K - 273.15) + 32 = K \): \[ \frac{9}{5}(K - 273.15) = K - 32 \] Multiply out: \[ 9K - 9 \times 273.15 = 5K - 160 \] Add like terms: \[ 4K = 2458.35 - 160 \] Solve for \( K \): \[ 4K = 2298.35 \] \[ K = 574.5875 \] Therefore, \( F = K = 574.59 \) (rounded to two decimal places).

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Key Concepts

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

Fahrenheit and Celsius
Fahrenheit and Celsius are two different scales used to measure temperature. You might wonder why we need different scales. Well, historically, different countries and scientists developed their methods of measuring temperature which led to different scales.

The Celsius scale is based on the freezing and boiling points of water. Water freezes at 0°C and boils at 100°C at sea level. This makes the Celsius scale easy to use and understand for everyday temperatures.

On the other hand, the Fahrenheit scale is a bit different. On this scale, water freezes at 32°F and boils at 212°F. It was developed by Daniel Fahrenheit in the early 18th century.

When converting between these two scales, the formula is:
  • Fahrenheit to Celsius: \( C = \frac{5}{9}(F - 32) \)
  • Celsius to Fahrenheit: \( F = \frac{9}{5}C + 32 \)
There is one temperature where Fahrenheit and Celsius agree with each other, which is \( -40 \, ^\circ \text{C} = -40 \, ^\circ \text{F}. \)
Fahrenheit and Kelvin
Let's talk about Fahrenheit and Kelvin. While Fahrenheit is common in the United States, Kelvin is used mostly in scientific contexts because it is an "absolute" scale. Here's why:

The Kelvin scale starts at absolute zero, the point where all molecular motion stops. This makes it handy for scientific purposes. In this scale, there are no negative numbers because Kelvin simply adds 273.15 to Celsius (\(K = C + 273.15\)). This means absolute zero is 0 K, equivalent to −273.15°C.

To convert from Fahrenheit to Kelvin, the formula is a bit more complex because it needs to account for the differences between the zero points and the scaling of each system:
  • Convert to Celsius:\( C = \frac{5}{9}(F - 32) \)
  • Then convert to Kelvin:\( K = C + 273.15 \)
Solving for when Fahrenheit equals Kelvin reveals one matching temperature, \( 574.59 \, K = 574.59 \, ^\circ \text{F} \)(rounded to two decimal places).
Temperature Conversion Formulas
Understanding temperature conversion formulas is crucial when dealing with multiple temperature scales. Let's break down the formulas you'll often use:

  • Celsius to Fahrenheit: \( F = \frac{9}{5}C + 32 \)
  • Fahrenheit to Celsius: \( C = \frac{5}{9}(F - 32) \)
  • Celsius to Kelvin: \( K = C + 273.15 \)
  • Kelvin to Celsius: \( C = K - 273.15 \)
  • Fahrenheit to Kelvin: First convert Fahrenheit to Celsius using \( C = \frac{5}{9}(F - 32) \), then use \( K = C + 273.15 \)
  • Kelvin to Fahrenheit: First convert to Celsius \( C = K - 273.15 \), then to Fahrenheit \( F = \frac{9}{5}C + 32 \)
By mastering these formulas, you can easily convert any temperature between different scales. Remembering these will improve your understanding of temperature relations and conversions.

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Most popular questions from this chapter

A Foucault pendulum consists of a brass sphere with a diameter of 35.0 cm suspended from a steel cable 10.5 m long (both measurements made at 20.0\(^\circ\)C). Due to a design oversight, the swinging sphere clears the floor by a distance of only 2.00 mm when the temperature is 20.0\(^\circ\)C. At what temperature will the sphere begin to brush the floor?

You have probably seen people jogging in extremely hot weather. There are good reasons not to do this! When jogging strenuously, an average runner of mass 68 kg and surface area 1.85 m\(^2\) produces energy at a rate of up to 1300 W, 80% of which is converted to heat. The jogger radiates heat but actually absorbs more from the hot air than he radiates away. At such high levels of activity, the skin's temperature can be elevated to around 33\(^\circ\)C instead of the usual 30\(^\circ\)C. (Ignore conduction, which would bring even more heat into his body.) The only way for the body to get rid of this extra heat is by evaporating water (sweating). (a) How much heat per second is produced just by the act of jogging? (b) How much net heat per second does the runner gain just from radiation if the air temperature is 40.0\(^\circ\)C (104\(^\circ\)F)? (Remember: He radiates out, but the environment radiates back in.) (c) What is the total amount of excess heat this runner's body must get rid of per second? (d) How much water must his body evaporate every minute due to his activity? The heat of vaporization of water at body temperature is \(2.42 \times 10{^6} J/kg\). (e) How many 750-mL bottles of water must he drink after (or preferably before!) jogging for a half hour? Recall that a liter of water has a mass of 1.0 kg.

The operating temperature of a tungsten filament in an incandescent light bulb is 2450 K, and its emissivity is 0.350. Find the surface area of the filament of a 150-W bulb if all the electrical energy consumed by the bulb is radiated by the filament as electromagnetic waves. (Only a fraction of the radiation appears as visible light.)

A 500.0-g chunk of an unknown metal, which has been in boiling water for several minutes, is quickly dropped into an insulating Styrofoam beaker containing 1.00 kg of water at room temperature (20.0\(^\circ\)C). After waiting and gently stirring for 5.00 minutes, you observe that the water’s temperature has reached a constant value of 22.0\(^\circ\)C. (a) Assuming that the Styrofoam absorbs a negligibly small amount of heat and that no heat was lost to the surroundings, what is the specific heat of the metal? (b) Which is more useful for storing thermal energy: this metal or an equal weight of water? Explain. (c) If the heat absorbed by the Styrofoam actually is not negligible, how would the specific heat you calculated in part (a) be in error? Would it be too large, too small, or still correct? Explain.

In a container of negligible mass, 0.200 kg of ice at an initial temperature of -40.0\(^\circ\)C is mixed with a mass m of water that has an initial temperature of 80.0\(^\circ\)C. No heat is lost to the surroundings. If the final temperature of the system is 28.0\(\circ\)C, what is the mass m of the water that was initially at 80.0\(^\circ\)C?

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