Lead is a soft, dense metal with a specific heat of 0.117 J/goC, a melting point of 328.0 oC, and a heat of fusion of 23012 J/g. How much heat (in kilojoules) must be provided to melt a 350.0 kg sample of lead that is initially at 20 oC?

q1 = heat required to raise T from 20 C to melting point.

q1 = (mass Pb(in g) x specific heat x (T m.p. - Tinitial)

q2 = heat required to melt the Pb at 328 but keep T at 328.
q2 = mass Pb x heat fusion.

Add q1 + q2 for q total.

To answer this question, we can calculate the heat required using the formula:

Q = m * (ΔT1 * c + ΔHfus + ΔT2 * c)

Where:
Q is the heat required (in joules)
m is the mass of the sample (in kilograms)
ΔT1 is the change in temperature from the initial temperature to the melting point of lead (in °C)
c is the specific heat of lead (in J/g°C)
ΔHfus is the heat of fusion of lead (in J/g)
ΔT2 is the change in temperature from the melting point to the final temperature (in °C)

First, let's convert the values provided into the appropriate units:
- The mass of the sample is given as 350.0 kg.
- The specific heat of lead is given as 0.117 J/goC. Since the specific heat is given in joules per gram-degree Celsius, we need to convert it to joules per kilogram-degree Celsius by dividing it by 1000, which gives us 0.117 J/g°C / 1000 = 0.000117 J/kg°C.
- The heat of fusion is given as 23012 J/g. Again, to convert it to joules per kilogram, we divide by 1000, which gives us 23.012 J/kg.

Next, we need to calculate the temperature changes:
- The initial temperature is given as 20 oC.
- The melting point of lead is given as 328.0 oC, so the change in temperature from the initial temperature to the melting point is: ΔT1 = 328.0 oC - 20 oC = 308.0 oC.
- Since the lead is being melted, the final temperature can be assumed to be the melting temperature, so the change in temperature from the melting point to the final temperature is ΔT2 = 328.0 oC - 328.0 oC = 0 oC.

Now we can substitute the values into the formula to calculate the heat required:

Q = 350.0 kg * (308.0 oC * 0.000117 J/kg°C + 23.012 J/kg + 0 oC * 0.000117 J/kg°C)
= 350.0 kg * (36.03636 J/kg + 23.012 J/kg)
= 350.0 kg * 59.04836 J/kg
= 20667.496 J

Finally, let's convert the result to kilojoules by dividing by 1000:
Q = 20667.496 J / 1000 = 20.667496 kJ

Therefore, the amount of heat required to melt a 350.0 kg sample of lead starting at 20 oC is approximately 20.67 kilojoules.