Answer:
24805.44 J
Step-by-step explanation:
Step 1:
Data obtained from the question.
Mass (M) = 464g
Initial temperature (T1) = 120°C
Final temperature (T2) = 219°C
Change in temperature (ΔT) = T2 - T1 = 219°C - 120°C = 99°C
Specific heat capacity of lead (C) = 0.129cal/g°C = 4.184 x 0.129 = 0.54J/g°C
Heat (Q) =?
Step 2:
Determination of the heat Q, required the temperature of lead. This is illustrated below
Q = MCΔT
Q = 464 x 0.54 x 99
Q = 24805.44 J
Therefore, 24805.44 J of heat is required to raise the temperature of lead.