If its initial temperature was 33.1°C, what is its final temperature?

Question

A block of aluminum with a volume of 99.0cm3 absorbs 65.2J of heat.  (density of aluminum =2.70 g/cm3)

Answer 1

Use the equation q = mC∆T

q = heat = 65.2 J

m = mass = 99.0 cm³ x 2.70 g / cm³ = 267.3 g

C = specific heat of aluminum = 0.921 J/gº (obtained from an online source)

∆T = change in temperature = ?

Solving for ∆T ...

65.2 J = (267.3 g)(0.921 J/gº)(∆T)

∆T = 0.2648º

Final temperature = 33.1º + 0.2648º = 33.36º = 33.4ºC (3 sig.figs.)