Question

An object with a mass of `15 g` is dropped into `250 mL` of water at `0^@C`. If the object cools by `120 ^@C` and the water warms by `9 ^@C`, what is the specific heat of the material that the object is made of?

Answer 1

From your data we can assume the object lost heat and the water gained heat in equilibrating.

Moreover, I am assuming:

`rho_(H_2O) = (1.0"g")/"mL"`

Recall,

`q = mC_sDeltaT`

In this case, the heat gained by the water is supplied by the body. Accordingly, we can rearrange the equation, such that,

`-m_xC_xDeltaT = m_(H_2O)C_(H_2O)DeltaT_(H_2O)`

Hence,

`=> C_x = (-m_(H_2O)C_(H_2O)DeltaT_(H_2O))/(mxDeltaT_x) approx (5.23"J")/("g"*°"C")`

is the specific heat of the body of mass we dropped into the water.