An object with a mass of #90 g# is dropped into #750 mL# of water at #0^@C#. If the object cools by #30 ^@C# and the water warms by #4 ^@C#, what is the specific heat of the material that the object is made of?

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Answer 1 · 2016-08-13T09:21:30

Given
#m_o->"Mass of the object"=90g#

#v_w->"Volume of water object"=750mL#

#Deltat_o->"Rise of temperature of water"=4^@C#

#Deltat_w->"Fall of temperature of the object"=30^@C#

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

#m_w->"Mass of water"#
#=v_wxxd_w=750mLxx1g/(mL)=750g#

#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#

#=>90xxs_o xx30=750xx1xx4#

#=>s_o=3000/2700=10/9#

#~~1.11calg^"-1"""^@C^-1#