A sample of aluminum absorbs 9.86 J of heat, upon which the temperature increases from 23.2°C to 30.5°C. Since the specific heat capacity of aluminum is 0.90 J/g-K, what is the mass of the sample?

1.50 g

Explanation:

The heat absorbed by the aluminum in this case is:

q = m x C x ΔT m = q / (C x ΔT)

q = 9.86 J

C = 0.90 J/g-K

ΔT = (30.5 ºC - 23.2 ºC) = 7.3 ºC = 7.3 K (this is a range of temperature)

m = 9.86 J / (0.90 J/g-K) x 7.3 K) = 1.50 g