# Question #b9477

May 5, 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 up has "m" grams of water. This mass of water absorbs 6345 J of heat. Its temperature increases from 22.6 °C ( 295.6 K) to 45.1 °C ( 318.1K)

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)

6345 J = m . 4190 J/kg∙K . ( 318.1 K - 295.6 K)

6345 J = m. 4190 J/kg∙K .22.5 K

6345 J = m 94275 J/ kg

m = 6345 J / 94275 J $k {g}^{- 1}$

m= 0.06730 kg or 673 g