heat= heataddedToIncreaseTemp+ heataddedToMelt
= 2kg*CPb*(327.5-25)+2kg*Hf
= 2kg*CPb*(327.5-25)+2kg*Hf
Q=78,650J Q=2.04x10^4 j/kg (2.00g)=40,800J
Q=119,450J „³ 119,000J
Q = m * �Hf
Where:
Q is the heat required
m is the mass of the lead (2.00 kg)
ï�„Hf is the heat of fusion for lead (2.04 × 10^4 J/kg)
Calculating Q:
Q = 2.00 kg * 2.04 × 10^4 J/kg
Q = 40,800 J
Therefore, you must transfer 40,800 J of heat to melt all the lead.
Let's break down the calculation step by step:
Step 1: Calculate the heat required to raise the temperature of the lead from 25.0°C to its melting point of 327.5°C.
The formula to calculate the heat required to raise the temperature of a substance is:
Q1 = m * Cp * ΔT
Where:
Q1 is the heat required
m is the mass of the substance (2.00 kg)
Cp is the specific heat capacity of the substance (130.0 J/kg°C)
ΔT is the change in temperature (327.5°C - 25.0°C)
Plugging in the values:
Q1 = 2.00 kg * 130.0 J/kg°C * (327.5°C - 25.0°C)
Step 2: Calculate the heat required to melt the lead.
The formula to calculate the heat required to melt a substance is:
Q2 = m * ΔHf
Where:
Q2 is the heat required
m is the mass of the substance (2.00 kg)
ΔHf is the heat of fusion for the substance (2.04 × 10^4 J/kg)
Plugging in the values:
Q2 = 2.00 kg * 2.04 × 10^4 J/kg
Step 3: Add both quantities of heat together to obtain the total heat required:
Total heat required = Q1 + Q2
Calculating Q1:
Q1 = 2.00 kg * 130.0 J/kg°C * (327.5°C - 25.0°C)
Q1 = 2.00 kg * 130.0 J/kg°C * 302.5°C
Calculating Q2:
Q2 = 2.00 kg * 2.04 × 10^4 J/kg
Adding Q1 and Q2:
Total heat required = Q1 + Q2
Finally, perform the calculations to get the total heat required.