Problem 28 - Chemistry: The Central Science

Question

(a) The temperature on a warm summer day is \(87^{\circ} \mathrm{F}\). What is the temperature in \({ }^{\circ} \mathrm{C}\) ? (b) Many scientific data are reported at \(25^{\circ} \mathrm{C}\). What is this temperature in kelvins and in degrees Fahrenheit? (c) Suppose that a recipe calls for an oven temperature of \(400^{\circ} \mathrm{F}\). Convert this temperature to degrees Celsius and to kelvins. (d) Liquid nitrogen boils at \(77 \mathrm{~K}\). Convert this temperature to degrees Fahrenheit and to degrees Celsius.

Step-by-Step Solution

Answer

(a) The temperature in Celsius is approximately \(30.56^{\circ} \mathrm{C}\). (b) The temperature is \(298.15 \mathrm{~K}\) and \(77^{\circ} \mathrm{F}\). (c) The temperature is \(204.44^{\circ} \mathrm{C}\) and \(477.59 \mathrm{~K}\). (d) The temperature is \(-196.15^{\circ} \mathrm{C}\) and \(-320.67^{\circ} \mathrm{F}\).

Step 1: Part (a): Fahrenheit to Celsius

To convert \(87^{\circ} \mathrm{F}\) to Celsius, we use the second formula: \(C = \frac{5}{9}(F - 32)\) Plug in the given temperature in Fahrenheit: \(C = \frac{5}{9}(87 - 32)\) Calculate the result: \(C \approx 30.56\) So the temperature in Celsius is approximately \(30.56^{\circ} \mathrm{C}\).

Step 2: Part (b): Celsius to Kelvin and Fahrenheit

We are given a temperature of \(25^{\circ} \mathrm{C}\). We will first convert it to Kelvin using the third formula: \(K = C + 273.15\) Plug in the given temperature in Celsius: \(K = 25 + 273.15\) Calculate the result: \(K = 298.15 \mathrm{~K}\) Now we will convert \(25^{\circ} \mathrm{C}\) to Fahrenheit using the first formula: \(F = \frac{9}{5}C + 32\) Plug in the given temperature in Celsius: \(F = \frac{9}{5}(25) + 32\) Calculate the result: \(F = 77^{\circ} \mathrm{F}\) So the temperature is \(298.15 \mathrm{~K}\) and \(77^{\circ} \mathrm{F}\).

Step 3: Part (c): Fahrenheit to Celsius and Kelvin

We need to convert \(400^{\circ} \mathrm{F}\) to Celsius and Kelvin. First, convert to Celsius using the second formula: \(C = \frac{5}{9}(F - 32)\) Plug in the given temperature in Fahrenheit: \(C = \frac{5}{9}(400 - 32)\) Calculate the result: \(C \approx 204.44^{\circ} \mathrm{C}\) Now, convert the Celsius result to Kelvin using the third formula: \(K = C + 273.15\) Plug in the temperature in Celsius: \(K = 204.44 + 273.15\) Calculate the result: \(K \approx 477.59 \mathrm{~K}\) So the temperature is \(204.44^{\circ} \mathrm{C}\) and \(477.59 \mathrm{~K}\).

Step 4: Part (d): Kelvin to Celsius and Fahrenheit

We need to convert \(77 \mathrm{~K}\) to Celsius and Fahrenheit. First, convert to Celsius using the fourth formula: \(C = K - 273.15\) Plug in the given temperature in Kelvin: \(C = 77 - 273.15\) Calculate the result: \(C \approx -196.15^{\circ} \mathrm{C}\) Now, convert the Celsius result to Fahrenheit using the first formula: \(F = \frac{9}{5}C + 32\) Plug in the temperature in Celsius: \(F = \frac{9}{5}(-196.15) + 32\) Calculate the result: \(F \approx -320.67^{\circ} \mathrm{F}\) So the temperature is \(-196.15^{\circ} \mathrm{C}\) and \(-320.67^{\circ} \mathrm{F}\).

Key Concepts

Celsius to Fahrenheit

Converting Celsius to Fahrenheit is a common temperature conversion used in everyday situations, such as understanding weather forecasts or cooking instructions. To perform this conversion, apply the formula \[ F = \frac{9}{5}C + 32 \]Where:

• \( F \) represents the temperature in degrees Fahrenheit
• \( C \) represents the temperature in degrees Celsius

The formula utilizes a conversion factor of \( \frac{9}{5} \) to shift between the two scales and adds 32 as an offset for alignment with the Fahrenheit system.
For example, if you have a temperature of \(25^{\circ} \mathrm{C} \) that you need to convert to Fahrenheit, you substitute the Celsius value in the equation:\[ F = \frac{9}{5}(25) + 32 \]Calculate to get:\[ F = 45 + 32 = 77^{\circ} \mathrm{F} \]Thus, \( 25^{\circ} \mathrm{C} \) converts to \( 77^{\circ} \mathrm{F} \).
This conversion is essential in settings like the United States, where Fahrenheit is more commonly used.

Fahrenheit to Celsius

To convert from Fahrenheit to Celsius, it's crucial to understand how the scales differ. Use the formula\[ C = \frac{5}{9}(F - 32) \]Here:

• \( C \) is the temperature in Celsius
• \( F \) is the temperature in Fahrenheit

This formula first reduces the Fahrenheit temperature by 32, the freezing point of water on this scale. Then, it adjusts by the factor \( \frac{5}{9} \), which reflects the relationship between the scales.
Consider a warm summer day temperature of \(87^{\circ} \mathrm{F} \). Substituting into the formula, you get:\[ C = \frac{5}{9}(87 - 32) \]\[ C \approx 30.56^{\circ} \mathrm{C} \]This calculation confirms that \(87^{\circ} \mathrm{F} \) is approximately \(30.56^{\circ} \mathrm{C} \).
Understanding this conversion is particularly useful in science and international travel, where Celsius is more commonly used.

Celsius to Kelvin

The Celsius and Kelvin scales are closely related, as both are part of the International System of Units (SI). This conversion is crucial in scientific contexts, as Kelvin is often used in physics and chemistry. The conversion is quite direct:\[ K = C + 273.15 \]Where:

• \( K \) stands for the temperature in Kelvin
• \( C \) is the temperature in Celsius

To convert \(25^{\circ} \mathrm{C} \) to Kelvin:\[ K = 25 + 273.15 \]\[ K = 298.15 \]So, \(25^{\circ} \mathrm{C} \) equals \(298.15 \mathrm{~K} \).
This conversion highlights that Kelvin does not have negative values, making it particularly useful for absolute temperature measurements in scientific studies.

Kelvin to Fahrenheit

Converting Kelvin to Fahrenheit involves a two-step process since Kelvin is not directly convertible to Fahrenheit. It first requires conversion from Kelvin to Celsius, and then to Fahrenheit.
Start by converting Kelvin to Celsius using:\[ C = K - 273.15 \]Subsequently, convert this result to Fahrenheit with:\[ F = \frac{9}{5}C + 32 \]For example, let's convert \(77 \mathrm{~K} \) to Fahrenheit:

• Convert Kelvin to Celsius: \( C = 77 - 273.15 \)
• Result: \( C = -196.15^{\circ} \mathrm{C} \)
• Convert Celsius to Fahrenheit: \( F = \frac{9}{5}(-196.15) + 32 \)
• Result: \( F \approx -320.67^{\circ} \mathrm{F} \)

Thus, \(77 \mathrm{~K} \) is approximately \(-320.67^{\circ} \mathrm{F} \).
Understanding this conversion is key for contexts involving cryogenics, where such low temperatures are common.