Lead has a melting point of 327ºC, its specific heat is J/g•deg, and its molar enthalpy of fusion is 4.80 kJ/mol. How much heat, in kilojoules, will be required to heat a 500.0-g sample of lead from 23.0ºC to its melting point and then melt it?

Ah, melting lead, the perfect way to make your cooking experiments go downhill! Let's do some calculations and have a laugh, shall we?

First, let's calculate the heat required to heat the lead from 23.0ºC to its melting point of 327ºC. We'll use the formula Q = mcΔT, where Q represents the heat, m is the mass, c is the specific heat, and ΔT is the change in temperature.

The specific heat of lead is missing in your question; please fill me in with the specific heat value!

To calculate the amount of heat required to heat the lead and then melt it, we need to consider two steps:

Step 1: Heating the lead from 23.0ºC to its melting point
Step 2: Melting the lead at its melting point

Step 1:
To heat the lead from 23.0ºC to its melting point, we need to calculate the heat using the equation:

q1 = m * C * ΔT

Where:
q1 = heat absorbed
m = mass of lead
C = specific heat of lead
ΔT = change in temperature

Given:
m = 500.0 g
C = specific heat of lead (not provided)
ΔT = (melting point of lead - initial temperature)

Since the specific heat of lead is not provided, we can assume it to be 0.128 J/g•ºC (specific heat of lead is approximately 0.128 J/g•ºC).

Substituting the values, we get:
ΔT = (327ºC - 23.0ºC) = 304ºC
C = 0.128 J/g•ºC

q1 = 500.0 g * 0.128 J/g•ºC * 304ºC
q1 = 19,712 J

Step 2:
To calculate the heat required to melt the lead at its melting point, we use the equation:

q2 = n * ΔHf

Where:
q2 = heat absorbed
n = number of moles of lead
ΔHf = molar enthalpy of fusion of lead

Given:
n = (mass of lead) / (molar mass of lead)
ΔHf = 4.80 kJ/mol

The molar mass of lead is 207.2 g/mol.

Substituting the values, we get:
n = 500.0 g / 207.2 g/mol
n = 2.41 mol

q2 = 2.41 mol * 4.80 kJ/mol
q2 = 11.57 kJ

Total Heat = q1 + q2
Total Heat = 19,712 J + 11.57 kJ
Total Heat = 31,282 J or 31.3 kJ

Therefore, the amount of heat required to heat a 500.0 g sample of lead from 23.0ºC to its melting point and then melt it is approximately 31.3 kJ.

To calculate the amount of heat required to heat and melt a sample of lead, we need to consider two steps: heating the lead from 23.0ºC to its melting point, and then melting it.

Step 1: Heating the lead from 23.0ºC to its melting point
To calculate the heat required to raise the temperature of lead from 23.0ºC to its melting point (327ºC), we can use the formula:

Q1 = mass × specific heat × change in temperature

Where:
Q1 is the heat required
mass is the mass of the lead sample (500.0 g)
specific heat is the specific heat capacity of lead (given in the problem)
change in temperature is the difference between the final temperature and initial temperature (327ºC - 23.0ºC)

Step 2: Melting the lead at its melting point
To calculate the heat required to melt the lead at its melting point, we can use the formula:

Q2 = moles × molar enthalpy of fusion

Where:
Q2 is the heat required
moles is the number of moles of lead
molar enthalpy of fusion is the enthalpy change per mole of substance during fusion (given in the problem)

First, let's calculate the number of moles of lead:
moles = mass ÷ molar mass

The molar mass of lead (Pb) is 207.2 g/mol. Therefore:
moles = 500.0 g ÷ 207.2 g/mol

Now, we can substitute the values into the formulas and calculate the heat required for each step.

Q1 = 500.0 g × specific heat × (327ºC - 23.0ºC)
Q2 = (500.0 g ÷ 207.2 g/mol) × 4.80 kJ/mol

Finally, add the calculated heats for both steps to find the total heat required:

Total heat required = Q1 + Q2

Substitute the values and calculate the final answer in kilojoules (kJ).

q1 = heat to raise T from 23.0C to 327 C.

q1 = mass Pb x specific heat Pb x (Tfinal-Tinitial)

q2 = heat to melt Pb
q2 = mass Pb x heat fusion.

Total = q1 + q2
I note you have mass in grams in specific heat (not listed) in J/g*C so that's ok. However, when you go to the melting stage, you have mass in grams and heat fusion in kJ/mol. I would change heat fusion to J/g*C. When you finish adding q1 and q2 the answer will be in joules, convet that to kJ.

31760J=31.76KJ