Question

If the melted lead got hot enough to vaporize, how much energy would it have taken to vaporize 21

Answer 1

Answer:

18 KJ

Explanation:

Data Given:

mass of Lead (m) = 21 g

Heat taken for vaporization (Q) = ?

Solution:

This problem is related to phase change and latent heat of vaporization.

Latent heat of vaporization is the amount of heat taken to convert one mole of substance at its boiling point to its vapor.

So, Latent heat of vaporization of lead has a constant value

Latent heat of vaporization of lead = 177.7 KJ/mol

Formula used

Q = m x Lv. . . . . (1)

where

Lv = specific latent heat of vaporization

here the value for latent heat of vaporization is for mole so instead of mass we will use moles in formula.

So,

Q = no. of mol x Lv. . . . . (2)

first find no. of moles for 21 g of lead

                  no. of moles = mass in grams / molar mass . . . . . . (3)

molar mass of  lead (Pb) = 207 g/mol

put values in equation 3

                      no. of moles = 21g / 207 g/mol

                      no. of moles = 0.101 mol

so,

number of moles of lead (Pb) = 0.101 mol

Put values in the eq.2

                      Q = 0.101 mol x 177.7 KJ/mol

                      Q = 18 kJ

So, 18KJ of heat is taken to vaporize 21 g of lead (Pb)