Answer:
B. 1,170,000 J
Explanation:
Given;
mass of lead block, m = 40 kg
initial temperature, t₁ = -25 ⁰C
final temperature, t₂ = 200 ⁰C
The heat absorbed the lead block is calculated as;
H = mcΔt
where;
c is the specific heat capacity of lead = 130 J/kg⁰C
H = 40 x 130 x (200 - (-25))
H = 40 x 130 x (200 + 25)
H = 40 x 130 x 225
H = 1,170,000 J
Therefore, the heat absorbed the lead block is 1,170,000 J
