Problem 5
Question
(a) At what temperature do the Fahrenheit and Celsius scales give the same reading? (b) Is there any temperature at which the Kelvin and Celsius scales coincide?
Step-by-Step Solution
Verified Answer
(a) -40 degrees is the temperature where Fahrenheit and Celsius are the same. (b) No temperature allows Kelvin and Celsius to have the same reading.
1Step 1: Understanding the Problem
The question asks us to find temperatures where two different temperature scales read the same value. First, we'll solve for where Fahrenheit equals Celsius, and then we'll determine if such a temperature exists for Kelvin and Celsius.
2Step 1: Equate Fahrenheit to Celsius
The relationship between Fahrenheit and Celsius is given by the formula: \[ F = \frac{9}{5}C + 32 \]We need to find the value of temperature 'T' where C = F. So, set C to F and solve for 'T'.\[ T = \frac{9}{5}T + 32 \]
3Step 2: Solve the Equation
To solve \( T = \frac{9}{5}T + 32 \), start by subtracting \( \frac{9}{5}T \) from both sides:\[ T - \frac{9}{5}T = 32 \]Simplify the left side by combining like terms:\[ \left(1 - \frac{9}{5}\right)T = 32 \]\[ -\frac{4}{5}T = 32 \]Multiply both sides by \(-\frac{5}{4}\) to isolate T:\[ T = -40 \]
4Step 3: Evaluate Kelvin and Celsius
The Kelvin scale is related to the Celsius scale by the formula:\[ K = C + 273.15 \]To find if a temperature exists where Kelvin equals Celsius, set K equal to C:\[ T = T + 273.15 \]This equation results in 0 = 273.15, which is not possible. Therefore, no temperature exists where Kelvin equals Celsius.
Key Concepts
Fahrenheit and Celsius equivalenceKelvin and Celsius differenceTemperature scale relationships
Fahrenheit and Celsius equivalence
Understanding when Fahrenheit and Celsius scales cross paths is a neat curiosity in temperature measurement. The formula to convert Celsius to Fahrenheit is \( F = \frac{9}{5}C + 32 \). To find a temperature where these scales are equal, we equate Celsius (C) to Fahrenheit (F). Thus, solving for \( T \), our equation becomes:
- Set \( C = F \)
- Resulting equation: \( T = \frac{9}{5}T + 32 \)
- \( \left(1 - \frac{9}{5}\right)T = 32 \)
- Simplifying yields \( -\frac{4}{5}T = 32 \)
Kelvin and Celsius difference
Kelvin and Celsius scales have a consistent difference and cannot be equivalent at any point. The relationship here is linear: \( K = C + 273.15 \). Kelvin is an absolute scale and always 273.15 units more than Celsius.
Setting these two scales to be equal results in the equation \( T = T + 273.15 \), which simplifies to \( 0 = 273.15 \), an impossibility. This shows that there isn't any temperature at which they coincide.
Setting these two scales to be equal results in the equation \( T = T + 273.15 \), which simplifies to \( 0 = 273.15 \), an impossibility. This shows that there isn't any temperature at which they coincide.
- Kelvin scale starts at absolute zero, with \( 0 \) Kelvin being the coldest possible temperature.
- Celsius scale starts at water’s freezing point, 0°C.
Temperature scale relationships
Understanding how different temperature scales relate provides a strong foundational grasp of their applications. Here's how the scales interrelate:
- Fahrenheit to Celsius: Utilize \( F = \frac{9}{5}C + 32 \). This indicates that Fahrenheit is an extended scale where water freezes at 32°F and boils at 212°F.
- Celsius to Kelvin: Simply shift by 273.15 using \( K = C + 273.15 \). This shows Kelvin is an absolute measure starting from absolute zero.
- The Kelvin scale is used primarily in scientific scenarios because it expresses temperature directly relative to absolute zero.
Other exercises in this chapter
Problem 3
(a) On January 22, 1943, the temperature in Spearfish, South Dakota, rose from \(-4.0^{\circ} \mathrm{F}\) to \(45.0^{\circ} \mathrm{F}\) in just 2 minutes. Wha
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Inside the earth and the sun. (a) Geophysicists have estimated that the temperature at the center of the earth's core is \(5000^{\circ} \mathrm{C}\) (or more),
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Convert the following Kelvin temperatures to the Celsius and Fahrenheit scales: (a) the midday temperature at the surface of the moon \((400 \mathrm{~K}) ;\) (b
View solution Problem 9
A metal rod is \(40.125 \mathrm{~cm}\) long at \(20.0^{\circ} \mathrm{C}\) and \(40.148 \mathrm{~cm}\) long at \(45.0^{\circ} \mathrm{C}\). Calculate the averag
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