Question

A 5.0g sample of aluminum with a specific heat of 0.90 j/g was heated from 22.1^c to 32.1^c. How much heat to the nearest joule did the aluminum gain?

Answer 1

the heat to the nearest joules that aluminium gain is calculated as below

Q(heat) =M(mass) x c ( specific heat capacity) x delta T( change in temperature)
M=5.0 g
c=0.90 j/g/c
deltat T= 32.1 -22.1=10c

q= 5.0 g x 0.90 j/g/c x 10 c= 45 joules is the one gained by Alluminium