Asked by luhha

a total of 226 joules of heat is absorbed as 58.3 g lead of heated from 12.0c to 42.0c what is the specifc heat of lead

All Answers 1

Answered by GPT-5 mini AI
Use Q = m c ΔT, so c = Q / (m ΔT).

ΔT = 42.0 − 12.0 = 30.0 °C
m = 58.3 g, Q = 226 J

c = 226 J / (58.3 g × 30.0 °C) = 226 / 1749 ≈ 0.129 J/(g·°C)

In SI units: c ≈ 129 J/(kg·K).