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Chapter 17: Problem 8

(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
(a) -40 degrees, (b) 576.09 Kelvin

Step by step solution

01

Understanding the Relationship Between Fahrenheit and Celsius

The relationship between Fahrenheit (F) and Celsius (C) is given by the conversion formula: \[ F = \frac{9}{5}C + 32 \]. To find the temperature at which Fahrenheit and Celsius are equal, we set \( F = C \).
02

Set Up the Equation for Fahrenheit and Celsius Agreement

Replace \( F \) with \( C \) in the equation: \[ C = \frac{9}{5}C + 32 \].
03

Solve for Celsius in the Fahrenheit-Celsius Equation

Subtract \( \frac{9}{5}C \) from both sides to get \[ C - \frac{9}{5}C = 32 \]. This simplifies to \[ -\frac{4}{5}C = 32 \]. Multiply both sides by \(-\frac{5}{4}\) to solve for \( C \): \[ C = -40 \].
04

Conclude the Agreement of Fahrenheit and Celsius

The temperature at which both Fahrenheit and Celsius thermometers read the same is \(-40\) degrees.
05

Understanding the Relationship Between Fahrenheit and Kelvin

The relationship between Fahrenheit (F) and Kelvin (K) can be established using Celsius as an intermediary: \( C = K - 273.15 \) and \( F = \frac{9}{5}C + 32 \). So we need \( F = K \).
06

Set Up the Equation for Fahrenheit and Kelvin Agreement

Substitute \( C = K - 273.15 \) in the Fahrenheit equation: \[ F = \frac{9}{5}(K - 273.15) + 32 \]. Simplify it to \( F = K \): \[ K = \frac{9}{5}(K - 273.15) + 32 \].
07

Solve for Kelvin in the Fahrenheit-Kelvin Equation

Distribute and simplify the equation: \[ K = \frac{9}{5}K - \frac{9}{5} \times 273.15 + 32 \]. This simplifies to: \[ K = \frac{9}{5}K - 491.67 + 32 \]. Rearrange to get: \[ 5K = 9K - 2304.35 \]. Simplifying further: \[ -4K = -2304.35 \]. Solve for \( K \): \[ K = 576.0875 \].
08

Conclude the Agreement of Fahrenheit and Kelvin

The temperature at which both Fahrenheit and Kelvin thermometers read the same is approximately \( 576.09 \) Kelvin.

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Key Concepts

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

Fahrenheit and Celsius relationship
Many people wonder about the point where Fahrenheit and Celsius temperature scales meet. It's fascinating because these two scales are widely used around the world. Each has its unique way of measuring temperature.
The relationship between these scales is governed by the formula:
  • \[ F = \frac{9}{5}C + 32 \]
This equation represents the conversion method from Celsius (\( C \)) to Fahrenheit (\( F \)). To find the temperature where both Fahrenheit and Celsius equal each other, you set \( F = C \).
By substituting in the equation, it simplifies to:
  • \[ -\frac{4}{5}C = 32 \]
From here, solving gives us \( C = -40 \).
Thus, at \-40 degrees, both Fahrenheit and Celsius thermometers agree.
Fahrenheit and Kelvin relationship
Have you ever wondered at which point Fahrenheit and Kelvin scales show the same number? Although Fahrenheit and Kelvin might seem unrelated, they can be connected through Celsius.
Here's how:
  • Start with the Celsius-Kelvin relationship: \( C = K - 273.15 \)
  • Utilize the Fahrenheit conversion formula \( F = \frac{9}{5}C + 32 \)
Now, for Fahrenheit and Kelvin to match, we set \( F = K \).
Substituting leads to:
  • \[ K = \frac{9}{5}(K - 273.15) + 32 \]
By solving this equation for \( K \), it simplifies down to \( K = 576.09 \).
Therefore, both Fahrenheit and Kelvin thermometers show the same temperature at approximately 576.09 Kelvin.
Solving temperature equations
Temperature equations can sometimes be a head-scratcher for students.
Understanding how to solve these equations is essential when moving between temperature scales like Fahrenheit, Celsius, and Kelvin.
  • Begin by learning the key conversion formulas among these scales.
  • Set the scale you want to equal another, like \( F = C \) or \( F = K \).
Apply algebraic manipulation, such as substituting and combining like terms, to isolate the variable you are solving for.
For example, solving \( K = \frac{9}{5}K - 491.67 + 32 \) involves rearranging and simplifying terms.
Being methodical with each step ensures accuracy and a deeper understanding of how these temperature conversion relationships work.

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Most popular questions from this chapter

"The Ship of the Desert" Camels require very little water because they are able to tolerate relatively large changes in their body temperature. While humans keep their body temperatures constant to within one or two Celsius degrees, a dehydrated camel permits its body temperature to drop to \(34.0^{\circ} \mathrm{C}\) overnight and rise to \(40.0^{\circ} \mathrm{C}\) during the day. To see how effective this mechanism is for saving water, calculate how many liters of water a 400 \(\mathrm{kg}\) camel would have to drink if it attempted to keep its body temperature at a constant \(34.0^{\circ} \mathrm{C}\) by evaporation of sweat during the day \(\left(12 \text { hours) instead of letting it rise to } 40.0^{\circ} \mathrm{C} \text { . (Note: The }\right.\) specific heat of a camel or other mammal is about the same as that of a typical human, 3480 \(\mathrm{J} / \mathrm{kg} \cdot \mathrm{K}\) . The heat of vaporization of water at \(34^{\circ} \mathrm{C}\) is \(2.42 \times 10^{6} \mathrm{J} / \mathrm{kg} .\) )

The tallest building in the world, according to some architectural standards, is the Taipei 101 in Taiwan, at a height of 1671 feet. Assume that this height was measured on a cool spring day when the temperature was \(15.5^{\circ} \mathrm{C}\) . You could use the building as a sort of giant thermometer on a hot summer day by carefully measuring its height. Suppose you do this and discover that the Taipei 101 is 0.471 foot taller than its official height. What is the temperature, assuming that the building is in thermal equilibrium with the air and that its entire frame is made of steel?

A crate of fruit with mass 35.0 \(\mathrm{kg}\) and specific heat 3650 \(\mathrm{J} / \mathrm{kg} \cdot \mathrm{K}\) slides down a ramp inclined at \(36.9^{\circ} \mathrm{C}\) below the horizontal. The ramp is 8.00 \(\mathrm{m}\) long. (a) If the crate was at rest at the top of the incline and has a speed of 2.50 \(\mathrm{m} / \mathrm{s}\) at the bottom, how much work was done on the crate by friction? (b) If an amount of heat equal to the magnitude of the work done by friction goes into the crate of fruit and the fruit reaches a uniform final temperature, what is its temperature change?

What is the rate of energy radiation per unit area of a black-body at a temperature of (a) 273 \(\mathrm{K}\) and \((\mathrm{b}) 2730 \mathrm{K} ?\)

Effect of a Window in a Door. A carpenter builds a solid wood door with dimensions \(2.00 \mathrm{m} \times 0.95 \mathrm{m} \times 5.0 \mathrm{cm} .\) Its thermal conductivity is \(k=0.120 \mathrm{W} / \mathrm{m} \cdot \mathrm{K}\) . The air films on the inner and outer surfaces of the door have the same combined thermal resistance as an additional \(1.8-\mathrm{cm}\) thickness of solid wood. The inside air temperature is \(20.0^{\circ} \mathrm{C}\) , and the outside air temperature is \(-8.0^{\circ} \mathrm{C} .\) (a) What is the rate of heat flow through the door? (b) By what factor is the heat flow increased if a window 0.500 \(\mathrm{m}\) on a side is inserted in the door? The glass is 0.450 \(\mathrm{cm}\) thick, and the glass has a thermal conductivity of 0.80 \(\mathrm{W} / \mathrm{m} \cdot \mathrm{K}\) . The air films on the two sides of the glass have a total thermal resistance that is the same as an additional 12.0 \(\mathrm{cm}\) of glass.
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