How much heat does 32.0 g of water absorb when it is heated from 25.0 to 80.0 °C?
1 Answer
May 10, 2017
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
