Problem 15 - Chemistry: Principles and Reactions

Question

Liquid helium is extensively used in research on superconductivity. Liquid helium has a boiling point of \(4.22 \mathrm{~K}\). Express this boiling point in \({ }^{\circ} \mathrm{C}\) and \(^{\circ} \mathrm{F}\).

Step-by-Step Solution

Answer

Answer: The boiling point of liquid helium in Celsius is -268.93°C, and in Fahrenheit, it is -452.07°F.

Step 1: Convert Kelvin to Celsius

To convert the given temperature from Kelvin to Celsius, we can use the following formula: \(T(^\circ C) = T(K) - 273.15\) Where: - \(T(^\circ C)\) is the temperature in Celsius - \(T(K)\) is the temperature in Kelvin In our case, \(T(K) = 4.22 \mathrm{~K}\). Plug this value into the formula and calculate the temperature in Celsius: \(T(^\circ C) = 4.22 - 273.15\)

Step 2: Calculate the boiling point in Celsius

Now let's calculate the boiling point in Celsius: \(T(^\circ C) = 4.22 - 273.15 = -268.93^\circ C\) So the boiling point of liquid helium in Celsius is \(-268.93^\circ C\).

Step 3: Convert Celsius to Fahrenheit

To convert the boiling point of liquid helium from Celsius to Fahrenheit, we'll use the following formula: \(T(^\circ F) = T(^\circ C) \times \frac{9}{5} + 32\) Where: - \(T(^\circ F)\) is the temperature in Fahrenheit - \(T(^\circ C)\) is the temperature in Celsius In our case, \(T(^\circ C) = -268.93^\circ C\). Plug this value into the formula and calculate the temperature in Fahrenheit: \(T(^\circ F) = -268.93 \times \frac{9}{5} + 32\)

Step 4: Calculate the boiling point in Fahrenheit

Now let's calculate the boiling point in Fahrenheit: \(T(^\circ F) = -268.93 \times \frac{9}{5} + 32 = -452.07^\circ F\) So the boiling point of liquid helium in Fahrenheit is \(-452.07^\circ F\). In conclusion, the boiling point of liquid helium in Celsius is \(-268.93^\circ C\) and in Fahrenheit is \(-452.07^\circ F\).

Key Concepts

Kelvin to Celsius Conversion

The Kelvin to Celsius conversion is a fundamental aspect of thermodynamics, especially in scenarios like scientific experiments. When working with temperatures in Kelvin, which is the SI unit of temperature, converting to Celsius can provide a clearer understanding for those more familiar with the Celsius scale.

The formula for conversion is straightforward: the temperature in Celsius, \(T(^\circ C)\), is the temperature in Kelvin, \(T(K)\), minus 273.15. For example, if the temperature is 4.22 Kelvin, we compute:

\[T(^\circ C) = T(K) - 273.15 = 4.22 - 273.15 = -268.93^\circ C\]

This shows us that at 4.22 Kelvin, which is the boiling point of liquid helium, the Celsius equivalent is \(-268.93^\circ C\). This temperature is significantly colder than any typical Earthly condition, illustrating the extremities in scientific experiments.

Celsius to Fahrenheit Conversion

To transition from Celsius to Fahrenheit, you apply a formula that scales the Celsius temperature and adjusts for the zero points of the two systems.

The formula for this conversion is:

\[T(^\circ F) = T(^\circ C) \times \frac{9}{5} + 32\]

When a temperature is given as \(-268.93^\circ C\), like the boiling point of liquid helium, we can convert it to Fahrenheit as follows:

\[-268.93 \times \frac{9}{5} + 32 = -452.07^\circ F\]

If you think that these numbers are extremely low, they are! This is because the Fahrenheit scale originates from a different basis, which makes it appear even colder than it already is in Celsius. Understanding this conversion is crucial, especially for applications spanning scientific and daily temperature assessments.

Superconductivity Research

Superconductivity is a mesmerizing field for both theoretical and applied physics. It refers to the phenomenon where a material can conduct electricity without resistance when it is below a certain critical temperature. This property becomes incredibly useful for applications requiring drastic reductions in energy dissipation.

Liquid helium, which has a boiling point at 4.22 Kelvin, is fundamental to superconductivity research because it is one of the only substances that can achieve such ultra-low temperatures. At these temperatures, many materials enter a superconductive state, which is pivotal for creating magnetic fields in scientific experiments and developing efficient power transmission solutions.

In research settings, liquid helium is often used to maintain the low temperatures needed for superconductor experiments. Without it, reaching the necessary conditions for superconductivity would be significantly more challenging. Through understanding how substances behave at such cold temperatures, scientists can explore new technologies that could revolutionize numerous industries, including medical imaging (like MRI machines), high-speed maglev trains, and quantum computing. These advancements underline the profound impact that low-temperature research and superconductivity can have on society.