Question

A 500.0 g sample of aluminium, I initially at 25.0 degrees, absorbs heat from its surroundings and reaches a final temperature of 90.7 degrees. How much heat (in KJ) has been absorbed by the sample? To one decimal place
Specific heat= 0.9930j g-1 K-1 for aluminium

Answer 1

Q= (500.0)(0.9930)(90.7 - 25.0)
Q= 32620.05