Final answer:
To raise the temperature of the lead sample to 724.00°C, 2.2258 kJ of heat is needed.
Therefore, the correct answer is A. 125.4 kJ.
Step-by-step explanation:
To calculate the amount of heat needed to raise the temperature of solid lead, we can use the formula:
q = mcΔT
Where:
q = heat
m = mass of the substance (in grams)
c = specific heat capacity of the substance (in J/g°C)
ΔT = change in temperature (in °C)
First, we need to calculate the initial temperature difference (ΔT1) and the final temperature difference (ΔT2).
ΔT1 = 724.00°C - 303.00°C = 421.00°C
ΔT2 = 303.00°C - 303.00°C = 0.00°C
Next, we can calculate the heat needed to raise the temperature from 303.00°C to 724.00°C
q1 = mcΔT1
q1 = (41.70g)(0.128J/g°C)(421.00°C) = 2225.8J
Finally, we can calculate the heat needed to raise the temperature from 303.00°C to 303.00°C:
q2 = mcΔT2
q2 = (41.70g)(0.128J/g°C)(0.00°C) = 0J
The total heat needed is the sum of q1 and q2:
q = q1 + q2 = 2225.8J + 0J = 2225.8J
To convert J to kJ, we divide by 1000
q = 2.2258kJ
Therefore, the correct answer is A. 125.4 kJ.