An object with a mass of #120 g# is dropped into #640 mL# of water at #0^@C#. If the object cools by #48 ^@C# and the water warms by #8 ^@C#, what is the specific heat of the material that the object is made of?

1 Answer
Mar 21, 2016

Answer:

#c_(obj)=4.465 J/(g""^oC) #

Explanation:

Given: An object with mass, #m_(obj) = 120g#
Object initial and final temperatures:
#T_(obj)^i=48^oC; T_(obj)^f=8^oC#
Water volume, #V_(H_2O)=640 mL, :. m_(H_2O)=640g#
Water initial and final temperature:
#T_w^i=0^oC; T_w^f=8^oC#
Physics Principle of Heat: Heat Loss from the Object is equal to Heat gained by water, at equilibrium.
#DeltaQ_(loss) = Q_("gained")#
Heat Loss#=>DeltaQ_(obj)= Q_f-Q_i = mc_(obj)(T_(obj)^f-T_(obj)^i)#
Heat Gain#=>DeltaQ_w= Q_f-Q_i = mc_w(T_w^f-T_w^i)#
#mc_o(T_(obj)^f-T_(obj)^i)=mc_w(T_w^f-T_w^i)# Substitute:
#120*color(red)(c_(obj))(48-8)=640*4.186(8-0) #
Solve for #color(red)(c_(obj))#
#c_(obj)=(64*cancel(8))/(12*cancel(40)^5)4.186=4.465 J/(g""^oC) #