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Chapter 19: Problem 33

what temperature do the following pairs of scales read the same, if ever: (a) Fahrenheit and Celsius (verify the listing in Table \(19-1\) ), (b) Fahrenheit and Kelvin, and (c) Celsius and Kelvin?

Short Answer

Expert verified
Celsius and Fahrenheit are equal at -40 degrees; Kelvin and Fahrenheit match at 574.585 K. No point exists where Celsius and Kelvin match in the same scale.

Step by step solution

01

Setup the equations

To find when two temperature scales read the same, set their formulas equal to each other. Use the following equations:1. Fahrenheit and Celsius: \( F = \frac{9}{5}C + 32 \).2. Fahrenheit and Kelvin: \( F = \frac{9}{5}(K - 273.15) + 32 \).3. Celsius and Kelvin: \( C = K - 273.15 \).
02

Solve the equations

Now solve the above equations for equality:(a) Fahrenheit and Celsius:Set \( F = C \)\( C = \frac{9}{5}C + 32 \);Reorganize to find the value of C that satisfies this equation.(b) Fahrenheit and Kelvin:Set \( F = K \)\( K = \frac{9}{5}(K - 273.15) + 32 \);Reorganize to find the value of K that satisfies this equation.(c) Celsius and Kelvin:Set \( C = K \)\(C = K - 273.15 \);Reorganize to find the value of C or K that satisfies this equation.
03

Calculate the temperatures

(a) Fahrenheit and Celsius:\( C = \frac{9}{5}C + 32 \);\( C - \frac{9}{5}C = 32 \);\( \frac{-4}{5}C = 32 \);\( C = -40 \);So at -40 degrees, Celsius and Fahrenheit are the same.(b) Fahrenheit and Kelvin:\( K = \frac{9}{5}(K - 273.15) + 32 \);Rearrange to \( K = 1.8 K - 491.67 + 32 \);\( K - 1.8 K = -459.67 \);\( -0.8 K = -459.67 \);\( K = 574.585 \);(c) Celsius and Kelvin:\( C = K - 273.15 \);Set C = K,\( K - 273.15 = K \);Not possible since this leads to no solution with finite temperature value where C = K in the common scale.

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Key Concepts

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

Fahrenheit and Celsius Equivalence
Understanding the point where Fahrenheit and Celsius scales read the same can be a bit tricky, but it's a fascinating concept. First, remember the formula to convert Celsius to Fahrenheit: ewline $$ F = \frac{9}{5}C + 32 $$To find the temperature where both scales equate, we set $$ F = C $$. This gives us:
  • $$ C = \frac{9}{5}C + 32 $$
By simplifying, you get:
  • $$ C - \frac{9}{5}C = 32 $$
  • $$\frac{-4}{5}C = 32 $$
  • $$ C = -40 $$
Thus, at -40 degrees, Celsius and Fahrenheit scales are the sameThis fact means (when you see -40) both mercury and alcohol thermometers will display the same reading. It鈥檚 a useful point especially in extremely cold climates.
Fahrenheit and Kelvin Equivalence
The Fahrenheit and Kelvin scales also have an intersection point. However, this one might seem less intuitive due to their origins. To find the common reading, use the conversion formula that involves Kelvin: ewline $$ F = \frac{9}{5}(K - 273.15) + 32 $$Set $$ F = K $$ and solve:
  • $$ K = \frac{9}{5}(K - 273.15) + 32 $$
  • Rearrange: $$ K = 1.8 K - 491.67 + 32 $$
  • Simplify: $$ K - 1.8 K = -459.67 $$
  • $$ -0.8 K = -459.67 $$
  • $$ K = 574.585 $$
Hence, at approximately 574.585 Kelvin, the Fahrenheit and Kelvin scales are equal.It's a significantly high value compared to regular physical phenomena, reflecting the different ideals these scales are based upon:
  • The Kelvin scale is absolute, starting from absolute zero.
  • The Fahrenheit scale is more conventional and practical in everyday usage.
Celsius and Kelvin Equivalence
When considering the Celsius and Kelvin scales, it鈥檚 important to note their direct relationship:ewline $$ C = K - 273.15 $$We naturally assume they might meet at a certain temperature. But substituting $$ C = K $$ reveals a contradiction:
  • $$ C = K - 273.15 $$
  • Set $$ C = K $$: $$ K = K - 273.15 $$
  • Simplifying, we see this equation leads to a false statement.
Therefore, Celsius and Kelvin scales never meet on a numerical plane.Yet, this intimacy helps in understanding both:
  • Kelvin essentially translates Celsius readings into absolute measurements.
  • Used often in scientific contexts because of its non-negative nature and starting from absolute zero.

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Most popular questions from this chapter

300 F Club You can join the semi-secret "300 F" club at the Amundsen-Scott South Pole Station only when the outside temperature is below \(-70^{\circ} \mathrm{C}\). On such a day, you first bask in a hot sauna and then run outside wearing only your shoes. (This is, of course, extremely dangerous, but the rite is effectively a protest against the constant danger of the winter cold at the south pole.) Assume that when you step out of the sauna, your skin temperature is \(102^{\circ} \mathrm{F}\) and the walls, ceiling, and floor of the sauna room have a temperature of \(30^{\circ} \mathrm{C}\). Estimate your surface area, and take your skin emissivity to be \(0.80 .\) (a) What is the approximate net rate \(P^{\text {net }}\) at which you lose energy via thermal radiation transfer to the room? Next, assume that when you are outside half your surface area transfers thermal radiation to the sky at a temperature of \(-25^{\circ} \mathrm{C}\) and the other half transfers thermal radiation to the snow and ground at a temperature of \(-80^{\circ} \mathrm{C}\). What is the approximate net rate at which you lose energy via thermal radiation exchanges with (b) the sky and (c) the snow and ground?

. Steam What mass of steam at \(100^{\circ} \mathrm{C}\) must be mixed with \(150 \mathrm{~g}\) of ice at its melting point, in a thermally insulated container, to produce liquid water at \(50^{\circ} \mathrm{C} ?\)

A \(1500 \mathrm{~kg}\) Buick moving at \(90 \mathrm{~km} / \mathrm{h}\) brakes to a stop, at a uniform rate and without skidding, over a distance of \(80 \mathrm{~m}\). At what average rate is mechanical energy transformed into thermal energy in the brake system?

At what temperature is the Fahrenheit scale reading equal to (a) twice that of the Celsius and (b) half that of the Celsius?

At \(20^{\circ} \mathrm{C}\), a rod is exactly \(20.05 \mathrm{~cm}\) long on a steel ruler. Both the rod and the ruler are placed in an oven at \(270^{\circ} \mathrm{C}\), where the rod now measures \(20.11 \mathrm{~cm}\) on the same ruler. What is the coefficient of thermal expansion for the material of which the rod is made?
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