To determine how much heat is released when 12.0 grams of helium gas condense, we will apply the formula related to latent heat:
[tex]\[ Q = mL \][/tex]
where:
- [tex]\( Q \)[/tex] is the heat released or absorbed during the phase change,
- [tex]\( m \)[/tex] is the mass of the substance,
- [tex]\( L \)[/tex] is the latent heat of vaporization.
Given data:
- The mass of helium ([tex]\( m \)[/tex]) is 12.0 grams.
- The latent heat of vaporization ([tex]\( L \)[/tex]) for helium is [tex]\( 21 \, \text{J/g} \)[/tex].
Step-by-Step Solution:
1. Identify the mass of helium:
[tex]\[ m = 12.0 \, \text{g} \][/tex]
2. Identify the latent heat of vaporization for helium:
[tex]\[ L = 21 \, \text{J/g} \][/tex]
3. Use the formula [tex]\( Q = mL \)[/tex] to calculate the heat released:
[tex]\[ Q = 12.0 \, \text{g} \times 21 \, \text{J/g} \][/tex]
4. Perform the multiplication:
[tex]\[ Q = 252.0 \, \text{J} \][/tex]
Therefore, the heat released when 12.0 grams of helium gas condense at 2.17 K is [tex]\( 252.0 \, \text{J} \)[/tex].
The correct answer among the provided choices is [tex]\( \boxed{250 \text{ J}} \)[/tex].
