An object with a mass of #32 g# is dropped into #480 mL# of water at #0^@C#. If the object cools by #120 ^@C# and the water warms by #4 ^@C#, what is the specific heat of the material that the object is made of?

1 Answer
Sep 23, 2016

Answer:

Specific heat of object is #0.50#.

Explanation:

Let the specific heat of object be #s#.

As quantum of heal gained / lost is given by #mxxsxx(t_2-t_1)#, where #m# is mass of object gaining / loosing heat and #s# is its specific heat and #(t_2-t_1)# is difference in temperature caused by heating / cooling,

heat lost by object is #32xxsxx120#

and heat gained by water is #480xx1xx4#.

Now heat lost should be equal to heat gained

#32xxsxx120=480xx1xx4#

i.e. #s=(480xx4)/(32xx120)=(cancel4xxcancel120xxcancel4)/(cancel4xxcancel4xx2xxcancel120)=1/3=0.50#

Hence specific heat of object is #0.50#.