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
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

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