Question

An object with a mass of `21 g` is dropped into `400 mL` of water at `0^@C`. If the object cools by `150 ^@C` and the water warms by `14 ^@C`, what is the specific heat of the material that the object is made of?

Answer 1

The heat energy lost by the object is the same as the energy gained by the water, or `Q_o=Q_w`

Explanation:

`c_o*m_o*DeltaT_o=c_w*m_w*DeltaT_w`

Now fill in what we know (`c_w=4.2` and `1mL=1g` water):
`c_o*21*150=4.2*400*14`

Work this out:
`c_o=(4.2*400*14)/(21*150)~~7.47`

This is a ridiculously high outcome, as no known material has a specific heat this high. Probably a `0` was lost or added in the problem.
With a mass of `210g` or an amount of water of `40mL` the outcome would be `0.747`, which is more plausible.