A 40.0 kg block of lead is heated from -25°C to 200°C. How much heat is absorbed by the lead block?

A. 2,354,000 J
B. 1,170,000 J
C. 56,891 J
D. 10,650 J

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

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