Which one of the following is the correct relation between Celsius and Fahrenheit temperature scales? (Symbols carry their usual meanings)
Understanding Celsius and Fahrenheit Temperature Relation
Temperature is a physical quantity that expresses the degree of hotness or coldness. It is measured using various scales, the most common being Celsius (\(\mathrm{T}_{\mathrm{C}}\)) and Fahrenheit (\(\mathrm{T}_{\mathrm{F}}\)). These scales relate to each other through a specific linear equation.
The Celsius scale is widely used globally, especially in scientific contexts and for everyday measurements in many countries. It defines the freezing point of water at standard atmospheric pressure as \(0^\circ\mathrm{C}\) and the boiling point as \(100^\circ\mathrm{C}\).
The Fahrenheit scale is primarily used in the United States and a few other countries. On this scale, the freezing point of water is defined as \(32^\circ\mathrm{F}\) and the boiling point as \(212^\circ\mathrm{F}\). The interval between the freezing and boiling points is \(212^\circ\mathrm{F} - 32^\circ\mathrm{F} = 180^\circ\mathrm{F}\).
Deriving the Celsius-Fahrenheit Formula
Since both scales are linear, we can establish a relationship between them based on their fixed points (freezing and boiling points of water). A change of \(100^\circ\mathrm{C}\) corresponds to a change of \(180^\circ\mathrm{F}\). This gives us a conversion factor:
- \(100^\circ\mathrm{C} \leftrightarrow 180^\circ\mathrm{F}\)
- \(1^\circ\mathrm{C} \leftrightarrow \frac{180}{100} ^\circ\mathrm{F} = \frac{9}{5} ^\circ\mathrm{F}\)
So, the slope of the conversion formula is \(\frac{9}{5}\). Now we need to account for the offset in the zero points. While \(0^\circ\mathrm{C}\) is the freezing point, the corresponding point in Fahrenheit is \(32^\circ\mathrm{F}\). Therefore, when \(\mathrm{T}_{\mathrm{C}} = 0\), \(\mathrm{T}_{\mathrm{F}}\) must be \(32\). This leads to the formula:
\(\mathrm{T}_{\mathrm{F}} = (\text{slope}) \times \mathrm{T}_{\mathrm{C}} + (\text{offset})\)
\(\mathrm{T}_{\mathrm{F}} = \frac{9}{5} \mathrm{T}_{\mathrm{C}} + 32\)
Analyzing the Given Options
Let's compare the derived correct relation with the provided options:
- \(\mathrm{T}_{\mathrm{F}}=\frac{10}{3} \mathrm{~T}_{\mathrm{C}}+32\) - Incorrect slope.
- \(\mathrm{T}_{\mathrm{F}}=\frac{5}{9} \mathrm{~T}_{\mathrm{C}}+36\) - Incorrect slope and offset. Note that \(\frac{5}{9}\) is for converting Fahrenheit to Celsius, not the other way around.
- \(\mathrm{T}_{\mathrm{F}}=\frac{9}{5} \mathrm{~T}_{\mathrm{C}}+36\) - Correct slope but incorrect offset.
- \(\mathrm{T}_{\mathrm{F}}=\frac{9}{5} \mathrm{~T}_{\mathrm{C}}+32\) - Correct slope and correct offset. This matches the standard formula relating Celsius and Fahrenheit.
Therefore, the correct relation between Celsius and Fahrenheit temperature scales is given by the formula \(\mathrm{T}_{\mathrm{F}}=\frac{9}{5} \mathrm{~T}_{\mathrm{C}}+32\).
Revision Table: Celsius vs. Fahrenheit Key Points
| Feature | Celsius (\(^\circ\mathrm{C}\)) | Fahrenheit (\(^\circ\mathrm{F}\)) |
|---|---|---|
| Freezing Point of Water | \(0^\circ\mathrm{C}\) | \(32^\circ\mathrm{F}\) |
| Boiling Point of Water | \(100^\circ\mathrm{C}\) | \(212^\circ\mathrm{F}\) |
| Interval between Freezing & Boiling | \(100^\circ\mathrm{C}\) | \(180^\circ\mathrm{F}\) |
| Conversion Factor (per \(^\circ\mathrm{C}\)) | \(1^\circ\mathrm{C}\) | \(\frac{9}{5}^\circ\mathrm{F}\) |
Additional Information: Other Temperature Scales
Besides Celsius and Fahrenheit, other temperature scales exist, notably the Kelvin (K) scale, which is the standard unit of thermodynamic temperature in the International System of Units (SI). The Kelvin scale is an absolute scale, meaning its zero point (0 K) is absolute zero, the theoretical point where particles have minimum vibration.
The relationship between Celsius and Kelvin is simple:
\(\mathrm{T}_{\mathrm{K}} = \mathrm{T}_{\mathrm{C}} + 273.15\)
There is no degree symbol used with Kelvin. \(0^\circ\mathrm{C}\) is equivalent to \(273.15\ \mathrm{K}\).
Another historical scale is the Reaumur scale (\(^\circ\mathrm{Re}\)), where the freezing point of water is \(0^\circ\mathrm{Re}\) and the boiling point is \(80^\circ\mathrm{Re}\).
Understanding these different scales and their relationships is crucial in various fields of physics and engineering.
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