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
Yonas Yohannes
Mar 21, 2016

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)

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