An object with a mass of 12 g12g is dropped into 240 mL240mL of water at 0^@C0∘C. If the object cools by 120 ^@C120∘C and the water warms by 6 ^@C6∘C, what is the specific heat of the material that the object is made of?

1 Answer
A08
Dec 6, 2016

0.5Cal//gm"^@C0.5Cal/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

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