An object with a mass of 120 g is dropped into 800 mL of water at 0^@C. If the object cools by 20 ^@C and the water warms by 3 ^@C, what is the specific heat of the material that the object is made of?

1 Answer
P dilip_k
Jul 23, 2016

1calg^-1C^-1

Explanation:

Given
m_o->"mass of the object"=120gm

v_w->"volume of water"=800mL

m_w->"vmass of water"=v_w xx "density of water"

=800mL xx1g/(mL)=800g

Deltat_w->"rise of temperature of water"=3^@C

Deltat_o->"fall of temperature of the object"=20^@C

s_w->"specific heat of water"=1 calg^-1C^-1

s_o->"specific heat of the bodyr"=?

By considering conservation of heat energy

"Heat lost by the object" ="Heat gained by water"

m_o xxs_o xxt_o=m_w xxs_w xxt_w

s_o =(m_w xxs_w xxt_w)/(m_o xxxxt_o)=(800xx1xx3)/(120xx20)

=1calg^-1C^-1

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