198k views
2 votes
2. A block of aluminum with a mass of 140g is cooled from 98.4oC to 62.2oC with a release of 1137J of heat. From these data, calculate the specific heat of aluminum.

User Sjeiti
by
6.8k points

1 Answer

3 votes
Q = M * C *ΔT

Q / ΔT = M

Δf - Δi = 98.4ºC - 62.2ºC = 36.2ºC

C = 1137 J / 140 * 36.2

C = 1137 / 5068

C = 0.224 J/gºC
User Daniel J F
by
6.7k points