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
Jul 23, 2016

Answer:

#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#