Calculate the amount of heat released when 24.8 g of water cools from 94° to 29°C. The specific heat of water is 4.184 J/g c

Answer 1

q = mC∆T

q = heat = ?

m = mass = 24.8 g

C = specific heat = 4.184 J/gº

∆T = change in temperature = 29º - 94º = -70º

q = (24.8 g)(4.184 J/gº)(-70º)

q = -7263 J = -7260 J (to 3 significant figures). The negative sign indicates heat is released.