Question & Answer: A 19.7 gram piece of lead at 123.5 °C is placed in a calorimeter containing water at 21.3 °C. If…..

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