A block of aluminum with a volume of 95.3 cm3 absorbs 65.7 J of heat. Part A If its initial temperature was 34.5 ∘C, what is its final temperature? (density of aluminum =2.70 g/cm3)

Answer 1

heat = mass x specific heat x change in temperature

q = m x C x ∆T

So, you'll need the specific heat for aluminum (from a table on the internet I found it to be 0.90 J/g/deg)

mass of aluminum = 95.3 cm3 x 2.70 g/cm3 = 257.3 g

65.7 J = (257.3 g)(0.90 J/g/deg)(∆T)

∆T = 0.28 degrees

Final temperature = 34.5º + 0.28º = 34.78º = 34.8ºC