how much heat in KJ must be added to raise the temp of a 300 g sample of water from 27.5 C to 39.1 C?

Answer 1

q = m x C x delta T
m = 300 g
C = 4.184 J/g°C
delta T = 39.1 - 27.5 = 11.6
q = 300 x 4.184 . 11.6 = 14,560 J = 14.56 kJ