Question

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?

Answer 1

`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`