Question

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?

Answer 1

`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)`