Question

An object with a mass of `32 g` is dropped into `250 mL` of water at `0^@C`. If the object cools by `60 ^@C` and the water warms by `3 ^@C`, what is the specific heat of the material that the object is made of?

Answer 1

Given
`m_o->"Mass of the object"=32g`

`v_w->"Volume of water object"=250mL`

`Deltat_w->"Rise of temperature of water"=3^@C`

`Deltat_o->"Fall of temperature of the object"=60^@C`

`d_w->"Density of water"=1g/(mL)`

`m_w->"Mass of water"`
`=v_wxxd_w=250mLxx1g/(mL)=250g`

`s_w->"Sp.heat of water"=1calg^"-1"""^@C^-1`

`"Let "s_o->"Sp.heat of the object"`

Now by calorimetric principle

Heat lost by object = Heat gained by water

`=>m_o xx s_o xxDeltat_o=m_wxxs_wxxDeltat_w`

`=>32xxs_o xx60=250xx1xx3`

`=>s_o=(250xx3)/(32xx60)`

`~~0.39calg^"-1"""^@C^-1`

Answer 2

masss of water= 250/1000= 1/4 kg
specific heat capacity of water= 42000J/`kg^0 C`
Change in temperature= `3^0 C`
Heat gained by water= mst = 1/442003= 3150
this is the heat lost by object
mass of object =32g =.032 kg
Change in temperature= `60^0 C`
Specific heat capacity of object = H/mt
=3150/.032*60
=1640.625 J / `kg^0 C`