An object with a mass of #120 g# is dropped into #720 mL# of water at #0^@C#. If the object cools by #24 ^@C# and the water warms by #3 ^@C#, what is the specific heat of the material that the object is made of?

1 Answer
Sep 19, 2016

Answer:

Specific heat of object is #0.75#.

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 objece is #120xxsxx24#

and heat gained by water is #720xx1xx3#.

Now heat lost should be equal to heat gained

#120xxsxx24=720xx1xx3#

i.e. #s=(720xx3)/(120xx24)=(6cancel720xxcancel3)/(cancel120xx8cancel24)=6/8=3/4=0.75#

Hence specific heat of object is #0.75#.