A sample of water at 75.6 C is cooled to 22.8 C. It released 19500 J of energy. What was the mass of water?

Answer 1

q = mC∆T

q = heat = -19500 J (- because heat was released, i.e. exothermic)

m = mass = ?

C = specific heat of water = 4.184 J/gº

∆T = change in temperature = 75.6 - 22.8 = -52.8º (- because temperature decreased)

solving for m, we have...

m = q / C∆T

m = -19500 J / (4.184 J/gº)(-52.8º)

m = 88.3 g