To determine how much heat is required to vaporize 15.0 kg of lead at its boiling point, we use the latent heat of vaporization for lead and the mass of lead given.
The formula to calculate the heat needed (Q) to vaporize a substance is:
[tex]\[ Q = m \cdot L_v \][/tex]
where:
- [tex]\( Q \)[/tex] is the heat required,
- [tex]\( m \)[/tex] is the mass of the substance,
- [tex]\( L_v \)[/tex] is the latent heat of vaporization.
Given:
- The latent heat of vaporization [tex]\(L_v\)[/tex] for lead is [tex]\( 8.68 \times 10^5 \, \text{J/kg} \)[/tex].
- The mass of lead [tex]\(m\)[/tex] is [tex]\( 15.0 \, \text{kg} \)[/tex].
By substituting these values into the formula, we get:
[tex]\[ Q = 15.0 \, \text{kg} \times 8.68 \times 10^5 \, \text{J/kg} \][/tex]
After performing the multiplication:
[tex]\[ Q = 1.302 \times 10^7 \, \text{J} \][/tex]
Therefore, the heat required to vaporize 15.0 kg of lead at its boiling point is:
[tex]\[ 1.30 \times 10^7 \, \text{J} \][/tex]
Hence, the correct answer is:
d. [tex]\(1.30 \times 10^7 \, \text{J}\)[/tex]
