Lead is a soft, dense metal with a specific heat of 0.028 kcal/kg·C°, a melting point of 328.0°C, and a heat of fusion of 5.5 kcal/kg. How much heat must be provided to melt a 202.2 kg sample of lead with a temperature of 20.0°C?

To calculate the heat required to melt a sample of lead, you can use the following formula:

Q = m × (ΔT × specific heat + heat of fusion)

Where:
Q = heat required (in kcal)
m = mass of the sample (in kg)
ΔT = change in temperature (in °C)
Specific heat = specific heat of lead (in kcal/kg·°C)
Heat of fusion = heat of fusion of lead (in kcal/kg)

Given:
Mass of the sample, m = 202.2 kg
Change in temperature, ΔT = (melting point of lead - initial temperature) = (328.0°C - 20.0°C) = 308.0°C
Specific heat of lead = 0.028 kcal/kg·°C
Heat of fusion of lead = 5.5 kcal/kg

Substituting the values into the formula, we get:

Q = 202.2 kg × (308.0°C × 0.028 kcal/kg·°C + 5.5 kcal/kg)

Calculating:

Q ≈ 202.2 kg × (8.624 kcal/kg + 5.5 kcal/kg)
Q ≈ 202.2 kg × 14.124 kcal/kg
Q ≈ 2856.1648 kcal

Therefore, approximately 2856.1648 kcal of heat must be provided to melt a 202.2 kg sample of lead with a temperature of 20.0°C.

To calculate the amount of heat required to melt a sample of lead, you need to consider the specific heat, the heat of fusion, and the temperature difference.

The specific heat of lead is given as 0.028 kcal/kg·C°. This value tells us the amount of heat required to raise the temperature of 1 kilogram of lead by 1 degree Celsius.

The heat of fusion of lead is given as 5.5 kcal/kg. This value tells us the amount of heat required to change the state of 1 kilogram of lead from solid to liquid at its melting point.

First, we need to find the temperature difference between the initial temperature (20.0°C) and the melting point of lead (328.0°C):
Temperature difference = melting point - initial temperature = 328.0°C - 20.0°C = 308.0°C

Next, we need to calculate the amount of heat required to raise the temperature of the lead from 20.0°C to its melting point using the specific heat formula:
Q1 = mass × specific heat × temperature difference
= 202.2 kg × 0.028 kcal/kg·C° × 308.0°C

Now, we calculate the amount of heat required to melt the lead at its melting point using the heat of fusion:
Q2 = mass × heat of fusion
= 202.2 kg × 5.5 kcal/kg

Finally, we add these two amounts of heat together to get the total heat required:
Total heat = Q1 + Q2

You can now substitute the given values into the equations and calculate the total heat required to melt the lead.

q1 = heat to raise temperature from 20 C to 328.

q1 = mass Pb x specific heat Pb x (328-20).

q2 = heat to melt Pb at 328.
q2 = mass Pb x delta H fusion.

Total heat = q1 + q2.

3531kcal