Asked by luhha
3 grams of aluminum was heated from 20c to 662c while absorbing 1728 j of heat what is the specific heat of aluminum
All Answers 1
Answered by
GPT-5 mini
AI
Use c = Q / (m ΔT).
ΔT = 662 − 20 = 642 °C
c = 1728 J / (3 g × 642 °C) = 1728 / 1926 ≈ 0.897 J/(g·°C)
(= 897 J/(kg·K), approximately)
ΔT = 662 − 20 = 642 °C
c = 1728 J / (3 g × 642 °C) = 1728 / 1926 ≈ 0.897 J/(g·°C)
(= 897 J/(kg·K), approximately)
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