An object with a mass of #12 g# is dropped into #240 mL# of water at #0^@C#. If the object cools by #120 ^@C# and the water warms by #6 ^@C#, what is the specific heat of the material that the object is made of?

1 Answer
Dec 6, 2016

Answer:

#0.5Cal//gm"^@C#

Explanation:

We know that in such interactions heat is gained by one and heat is lost is by the other.

Also that the heat gained/lost is given by
#DeltaQ=msDeltat#
where #m,s and Deltat# are the mass, specific heat and rise or gain in temperature of the object
#Delta Q_"lost"=Delta Q_"gained"#

In the given problem heat is lost by the object and gained by water at #0^@C#. Using CGS system of units, where #1mL# of water has a mass of #1gm#

Heat gained by water
#DeltaQ_"gained"=msDeltat#
#DeltaQ_"gained"=240xx1xx6=1440Cal# ......(1)

And heat lost by object whose specific heat is #s_o# is given by
#Delta Q_"lost"=m_"object"cdot s_ocdot Deltat#
#=>Delta Q_"lost"=12cdot s_ocdot120#
#=>Delta Q_"lost"=1440 s_o# ......(2)
Equating (1) and (2) and solving for the required quantity
#1440=1440 s_o#
#=>s_o=1440/1440=1Cal//gm^@C#