Asked by qp
The initial temperature of a bomb calorimeter is 28.50°C. When a chemist carries out a reaction in this calorimeter, its temperature decreases to 27.45°C. If the calorimeter has a mass of 1.400 kg and a specific heat of 3.52 J/(gi°C), how much heat is absorbed by the reaction?
Use q equals m C subscript p Delta T..
140 J
418 J
1,470 J
5,170 J
All Answers 1
Answered by
GPT-5 mini
AI
First compute the calorimeter's heat change: q_cal = m C ΔT.
Convert mass to grams: 1.400 kg = 1400 g.
ΔT = 27.45 − 28.50 = −1.05 °C.
q_cal = (1400 g)(3.52 J/g·°C)(−1.05 °C) = −5174 J (≈ −5.17 × 10^3 J).
The reaction absorbs the heat lost by the calorimeter, so q_reaction = +5174 J ≈ 5,170 J.
Answer: 5,170 J.
Convert mass to grams: 1.400 kg = 1400 g.
ΔT = 27.45 − 28.50 = −1.05 °C.
q_cal = (1400 g)(3.52 J/g·°C)(−1.05 °C) = −5174 J (≈ −5.17 × 10^3 J).
The reaction absorbs the heat lost by the calorimeter, so q_reaction = +5174 J ≈ 5,170 J.
Answer: 5,170 J.
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