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?
2 Answers
Given
Now by calorimetric principle
Heat lost by object = Heat gained by water
masss of water= 250/1000= 1/4 kg
specific heat capacity of water= 42000J/
Change in temperature=
Heat gained by water= mst = 1/442003= 3150
this is the heat lost by object
mass of object =32g =.032 kg
Change in temperature=
Specific heat capacity of object = H/mt
=3150/.032*60
=1640.625 J /