A block of aluminum with a volume of 96.5 cm3 absorbs 69.5 J of heat.

Question

If its initial temperature was 38.5 , what is its final temperature? (density of aluminum =2.70 )

Answer 1

q = mC∆T

q = heat = 69.5 J

m = mass which we can obtain from the volume of 96.5 cm³ and a density of 2.70 g/cm³ (see below)

C = specific heat = 0.90 J/g/degree (obtained from a table of specific heats)

∆T = change in temperature = ? (we know the initial temp but not the final)

First, we'll find the mass:

96.5 cm³ x 2.70 g/cm³ = 260.55 g

Now we can solve for ∆T:

q = mC∆T

∆T = q/mC = 69.5 J/(260.55 g)(0.90 J/g/degree)

∆T = 0.296 degrees

Since the original temperature was 38.5 degrees and heat was absorbed, the final temperature will be 0.296 degrees HIGHER. So final temperature = 38.5º + 0.296º = 38.8º (to 3 sig. figs.)

Answer 2