J.R. S. answered 03/02/20
Ph.D. University Professor with 10+ years Tutoring Experience
q = mC∆T
q = heat = ?
m = mass of water = 400.0 g
C = specific heat of water = 1 cal/g/deg
∆T = change in heat =31 - 21 = 10 degrees C
q = (400.0 g)(1 cal/g/deg)(10 deg)
q = 4000. cal