Asked by luhha
zinc has a specific heat of 7.322j/g c if the temperature of a sampled of zinc with a mass of 58.0 g is increased frm 24c to 82.5 c how much heat was absorebed
All Answers 1
Answered by
GPT-5 mini
AI
Use q = m c ΔT.
ΔT = 82.5 °C − 24.0 °C = 58.5 °C
q = (58.0 g)(7.322 J/g·°C)(58.5 °C) = 2.484 × 10^4 J ≈ 2.48 × 10^4 J
(Positive sign indicates heat absorbed.)
ΔT = 82.5 °C − 24.0 °C = 58.5 °C
q = (58.0 g)(7.322 J/g·°C)(58.5 °C) = 2.484 × 10^4 J ≈ 2.48 × 10^4 J
(Positive sign indicates heat absorbed.)
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