How much heat does 32.0 g of water absorb when it is heated from 25.0 to 80.0 °C?
We can answer this question using specific heat concept. Water has to absorb 4.184 Joules of heat for the temperature of one gram of water to increase by 1 degree celsius (°C), its Specific heat is
4.184 J/g°C or 4190 J/kg∙K
let us assume that the 32 grams of water. This mass of water absorbs Q J of heat. Its temperature increases from 25 °C (298K) to 80°C ( 353 K)
applying Q = mc∆T
Q = heat energy (Joules, J)
m = mass of a substance (kg)
c = specific heat (units J/kg∙K)
∆ is a symbol meaning "the change in"
∆T = change in temperature (Kelvins, K)
Q = 0.032kg . 4190 J/kg∙K . ( 353 K - 298 K)
Q = 0.032 kg . 4190 J/kg∙K . 55 K
Q = 7374 J