An object with a mass of #32 g# is dropped into #250 mL# of water at #0^@C#. If the object cools by #16 ^@C# and the water warms by #3 ^@C#, what is the specific heat of the material that the object is made of?

Question

999 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-11T08:59:41

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"=16^@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 xx16=250xx1xx3#

#=>s_o=(250xx3)/(32xx16)#

#~~1.46calg^"-1"""^@C^-1#