How much heat is absorbed when a 35.0 g aluminum block increases in temperature from 23°C to 33°C?

Specific heat of Al: 0.900 J*g-1*°C-1

q=msΔT
q=(35.0 g)(.900)(33°C - 23°C)

q= 315 J? Just making sure there isn't anything that I am missing.

1 answer

I didn't check the math on the calculator but the process and the numbers are correct.