A 168 g aluminum engine part at an initial temperature of 16.0°C absorbs 1,949 J of heat. What is the final temperature of the part (°C)?
(c of Al = 0.900 J/g·K)?
E = m c ΔT
1949 J = (168 g) × [0.900 J/(g °C)] × ΔT
ΔT = [1949 / (168 × 0.900)]°C = 12.9°C
Final temperature = (16.0 + 12.9)°C = 28.9°C
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E = m c ΔT
1949 J = (168 g) × [0.900 J/(g °C)] × ΔT
ΔT = [1949 / (168 × 0.900)]°C = 12.9°C
Final temperature = (16.0 + 12.9)°C = 28.9°C
https://answers.yahoo.com/question/index?qid=20081...