An object with a mass of 15 g is dropped into 250 mL of water at 0^@C. If the object cools by 120 ^@C and the water warms by 2 ^@C, what is the specific heat of the material that the object is made of?

1 Answer
Aug 11, 2016

~~0.28calg^"-1"""^@C^-1

Explanation:

Given
m_o->"Mass of the object"=15g

v_w->"Volume of water object"=250mL

Deltat_w->"Rise of temperature of water"=2^@C

Deltat_o->"Fall of temperature of the object"=120^@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

=>15xxs_o xx120=250xx1xx2

=>s_o=(250xx2)/(15xx120)=5/18

~~0.28calg^"-1"""^@C^-1