q = mass Al x specific heat Al x (Tfinal-Tinitial)
85 = 295 x 0.900 x (Tfinal - 3.00)
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)?
1 answer