If a 375 mL sample of water was cooled from 37.5 C to 0 C how much heat wad lost in joules

1 mL = 1 gram

Q=375*4.179*37.5

The equation for heat is

Q = m*Cp*del (T) where

Q = heat in Joules

m = mass in kg

Cp = the heat capacity of water at a constant pressure = 4,184 J/kgC

del (T) = the change in temperature in either K or degrees C

1 liter of water = 1 kg at standard temperature and pressure.

375 mL water x (1 kg/1L water) = 0.375 kg water.

Q = (0.375 kg water) (4184 J/kgC) (37.5 C - 0 C) = 58,800 Joules.