Asked by Anonymous
A 295-g aluminum engine part at an initial temperature of3.00°C absorbs 85.0 kJ of heat.
What is the final temperature ofthe part (specific heat of Al = 0.900 J/g·°C)?
What is the final temperature ofthe part (specific heat of Al = 0.900 J/g·°C)?
Answers
Answered by
DrBob222
q = mass Al x specific heat Al x (Tfinal-Tinitial)
85 = 295 x 0.900 x (Tfinal - 3.00)
85 = 295 x 0.900 x (Tfinal - 3.00)
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