Question

Question #b9477

Answer 1

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 `kg^(-1)`

m= 0.06730 kg or 673 g