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
P dilip_k
Aug 18, 2016

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#

Mahendra S.
Aug 18, 2016

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#

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