Question

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

Answer 1

1 mL = 1 gram
Q=375*4.179*37.5

Answer 2

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.