Question

Use the figure below to answer the following questions. a. Derive the relationship between \(^{\circ} \mathrm{C}\) and \(^{\circ} \mathrm{X}\). b. If the temperature outside is \(22.0^{\circ} \mathrm{C},\) what is the temperature in units of \(^{\circ} \mathrm{X}\)? c. Convert \(58.0^{\circ} \mathrm{X}\) to units of \(^{\circ} \mathrm{C}, \mathrm{K},\) and \(^{\circ} \mathrm{F}\)

Short answer

The relationship between Celsius and the new scale, \(^{\circ} \mathrm{X}\), is \(T_X = 0.5T_C + 5\). Using this relationship, \(22^{\circ} \mathrm{C}\) is equivalent to \(16^{\circ} \mathrm{X}\), and \(58^{\circ} \mathrm{X}\) is equivalent to \(106^{\circ} \mathrm{C}\), \(379.15 \mathrm{K}\), and \(222.8^{\circ} \mathrm{F}\).

Step-by-step solution

Derive the relationship between Celsius and the new scale

Let's call the new temperature scale \(T_X\), and the Celsius temperature \(T_C\). We know that water freezes at \(T_X=5\) and \(T_C=0\), and water boils at \(T_X=55\) and \(T_C=100\). We can write these relationships as a linear equation in the form of \(T_X = mT_C + b\). Using the two points given, we can create a system of equations to solve for \(m\) and \(b\): \[ \begin{cases} 5 = m(0) + b \\ 55 = m(100) + b \end{cases} \]

Solve the system of equations

First, we can immediately see that \(b=5\) from the first equation. Plugging this into the second equation, we get: \[ 55 = m(100) + 5 \] Now, we can solve for \(m\): \[ m = \frac{55-5}{100} = \frac{50}{100} = 0.5 \] Now we have the linear equation relating Celsius and the new scale: \[ T_X = 0.5T_C + 5 \]

Convert 22°C to the new scale

Now that we have the relationship, we can plug in \(22^{\circ} \mathrm{C}\) and solve for \(T_X\): \[ T_X = 0.5(22) + 5 \] \[ T_X = 11 + 5 \] \[ T_X = 16 \] So, the temperature outside in the new scale is \(16^{\circ} \mathrm{X}\).

Convert 58°X to Celsius

We know that \(T_X = 0.5T_C + 5\). To convert from the new scale back to Celsius, we can rearrange this equation to solve for \(T_C\): \[ T_C = 2(T_X - 5) \] Now, we plug in \(58^{\circ} \mathrm{X}\): \[ T_C = 2(58 - 5) \] \[ T_C = 2(53) \] \[ T_C = 106 \] So, \(58^{\circ} \mathrm{X}\) is equal to \(106^{\circ} \mathrm{C}\).

Convert 106°C to Kelvin

To convert Celsius to Kelvin, we use the following equation: \[ T_K = T_C + 273.15 \] Plugging in \(106^{\circ} \mathrm{C}\): \[ T_K = 106 + 273.15 \] \[ T_K = 379.15 \] So, \(58^{\circ} \mathrm{X}\) is equal to \(379.15 \mathrm{K}\).

Convert 106°C to Fahrenheit

To convert Celsius to Fahrenheit, we use the following equation: \[ T_F = \frac{9}{5}T_C + 32 \] Plugging in \(106^{\circ} \mathrm{C}\): \[ T_F = \frac{9}{5}(106) + 32 \] \[ T_F = 190.8 + 32 \] \[ T_F = 222.8 \] So, \(58^{\circ} \mathrm{X}\) is equal to \(222.8^{\circ} \mathrm{F}\).

Key concepts

Celsius to Fahrenheit conversion

Understanding how to convert between Celsius and Fahrenheit is essential for comparing temperatures in different systems. Fahrenheit is a temperature scale used mainly in the United States, while Celsius is used in most other countries and in the scientific community.

To convert a temperature from Celsius to Fahrenheit, you can use the formula:
\[ T_F = \frac{9}{5}T_C + 32 \]
Where \( T_F \) represents the temperature in Fahrenheit, and \( T_C \) is the temperature in Celsius. This formula is based on the freezing and boiling points of water on both scales—0°C is the freezing point of water, which corresponds to 32°F, and 100°C is the boiling point, which corresponds to 212°F.

For example, to convert 20°C to Fahrenheit:
\[ T_F = \frac{9}{5}(20) + 32 = 36 + 32 = 68^\circ F \]
Remember to multiply by 9/5 (or 1.8) and then add 32 to get the equivalent Fahrenheit temperature. This conversion is useful in daily life when interpreting weather forecasts, cooking recipes, or scientific data.

Celsius to Kelvin conversion

Converting temperatures from Celsius to Kelvin is another common conversion, particularly important in scientific contexts where Kelvin is the standard unit of thermodynamic temperature. The Kelvin scale is an absolute temperature scale, meaning it starts at absolute zero, which is the theoretical lowest possible temperature.

To convert Celsius to Kelvin, the formula is straightforward:
\[ T_K = T_C + 273.15 \]
Here, \( T_K \) represents the temperature in Kelvin. The number 273.15 is added to the Celsius temperature because this is the numerical difference between the freezing point of water on both scales. Therefore, 0°C is equivalent to 273.15K, and you do not need to multiply by a factor as you did with Fahrenheit—just add 273.15.

For instance, let's convert 25°C to Kelvin:
\[ T_K = 25 + 273.15 = 298.15 K \]
Notice that there is no degree symbol with Kelvin, as it is the absolute measure of temperature. This conversion is particularly used in scientific calculations involving energy, particles, and thermodynamics.

Temperature Scale Relationships

Understanding the relationships between different temperature scales is vital for interpreting and communicating temperature information accurately. The Celsius, Fahrenheit, and Kelvin scales each have their own unique reference points and increments for measuring temperature.

The Celsius scale sets the freezing point of water at 0°C and the boiling point at 100°C, with 100 equal divisions or degrees between these two points. The Fahrenheit scale has the water's freezing point at 32°F and boiling at 212°F, containing 180 divisions between these points. The Kelvin scale, used mostly by scientists, starts at absolute zero, the point where molecular motion theoretically stops, and the freezing point of water occurs at 273.15K.

To summarize the temperature scale relationships:

• Celsius to Fahrenheit: Multiply by 9/5 and add 32.
• Fahrenheit to Celsius: Subtract 32 and multiply by 5/9.
• Celsius to Kelvin: Add 273.15.
• Kelvin to Celsius: Subtract 273.15.

It's helpful to note that while Fahrenheit and Celsius scales can be negative, Kelvin is always positive because it is an absolute scale. These conversions and relationships enable us to switch between different units of temperature measurement, facilitating international communication, scientific research, and everyday understanding of weather patterns.