A 19.7 gram piece of lead at 123.5 °C is placed in a calorimeter containing water at 21.3 °C. If the temperature at equilibrium is 24.5 °C, what is the mass of the water?
Expert Answer
Assume only water absorbs energy and the colorimeter won’t
so
-Qlost = Qgain
-Qmetal = Qwater
Qmetal = mmetal * Cmetal * (Tf-Tmetal)
Qwater = mwater * Cwater * (Tf – Twater)
substitte known data
Clead 0.13 J/gC, Cwater = 4.184 J/gC
the euqation
– mmetal * Cmetal * (Tf-Tmetal) = mwater * Cwater * (Tf – Twater)
substitute
-19.7*0.13*(24.5 – 123.5) = mwater * 4.184*(24.5-21.3)
253.539 = 13.3888*mwater
mwater = 253.539/13.3888
mwater = 18.93 g