Help me understand this please
You have a 2.00 kg block of lead. Lead melts at 327.5°C. CPb = 130.0 , and Hf for lead is 2.04 × 10^4 J/kg. Say you start at room temperature (25.0°C). How much heat must you transfer to melt all the lead?
To find out how much heat must be transferred to melt the lead, we need to calculate the heat required to raise the temperature of the lead from room temperature to its melting point, and then the heat required for the phase change from solid to liquid.
Step 1: Calculate the heat required to raise the temperature from 25.0°C to its melting point:
The formula to calculate heat (Q) is given by:
Q = mcΔT
Where:
Q is the heat required
m is the mass of the substance
c is the specific heat capacity of the substance
ΔT is the change in temperature
Given:
Mass of lead (m) = 2.00 kg
Specific heat capacity of lead (cPb) = 130.0 J/kg·°C
Change in temperature (ΔT) = Melting point of lead (327.5°C) - Room temperature (25.0°C)
Using the formula above, we can calculate the heat required to raise the temperature:
Q1 = (2.00 kg) × (130.0 J/kg·°C) × (327.5°C - 25.0°C)
Step 2: Calculate the heat required for the phase change:
The heat required for the phase change is given by the formula:
Q2 = m × ΔHf
Where:
Q2 is the heat required for the phase change
m is the mass of the substance
ΔHf is the latent heat of fusion
Given:
Mass of lead (m) = 2.00 kg
ΔHf for lead = 2.04 × 10^4 J/kg
Using the formula above, we can calculate the heat required for the phase change:
Q2 = (2.00 kg) × (2.04 × 10^4 J/kg)
Step 3: Calculate the total heat required:
The total heat required is the sum of Q1 and Q2:
Total heat required = Q1 + Q2
Now you can substitute the values into the equation to find the answer.