Find the energy needed to convert 28  kg of liquid oxygen at its boiling point into gas.
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To find the energy required to convert 28 kg of liquid oxygen (O₂) at its boiling point into gas, we need to use the latent heat of vaporization formula:

\[
Q = m \cdot L
\]

where:
- \( Q \) is the energy required (in joules),
- \( m \) is the mass of the substance (in kilograms),
- \( L \) is the latent heat of vaporization (in joules per kilogram).

For oxygen (O₂), the latent heat of vaporization is approximately:

\[
L = 213 \, \text{kJ/kg} = 213,000 \, \text{J/kg}
\]

Now, we can calculate:

\[
Q = 28 \, \text{kg} \cdot 213,000 \, \text{J/kg} = 5,964,000 \, \text{J}
\]

So, the energy required is \( 5,964,000 \, \text{J} \) or approximately \( 5.96 \, \text{MJ} \).

Would you like more details or have any questions?

Here are 5 related questions for further exploration:
1. What is the latent heat of fusion for oxygen?
2. How does the boiling point of oxygen compare to other gases?
3. What is the specific heat capacity of oxygen gas?
4. How would you calculate the energy for a different mass of oxygen?
5. What role does pressure play in the boiling point of oxygen?

**Tip:** Always ensure units are consistent in thermodynamics problems to avoid errors in energy calculations!

Find the energy needed to convert 28 kg of liquid oxygen at its boiling point into gas.

Discover how to calculate the energy needed to convert 28 kg of liquid oxygen at its boiling point into gas using the latent heat of vaporization formula. This thermodynamics problem involves key concepts such as latent heat and energy conversion, suitable for high school and undergraduate students.

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