Question

An object with a mass of `125 g` is dropped into `750 mL` of water at `0^@C`. If the object cools by `40 ^@C` and the water warms by `3 ^@C`, what is the specific heat of the material that the object is made of?

Answer 1

`c_{object}=1.88J/{g\quad^o C}=0.45\text{cal}/{g\quad^o C}`

Explanation:

Because energy is conserved, the heat gained by the water equals the heat lost by the object.

`Q_{object}=-Q_{water}`

The minus sign is there because one object is losing heat while the other is absorbing heat.

The specific heat of water is known, `\quad c_{water}=4.18 J/{g\quad^o C}`

1 mL of water is the same as 1 gram of water so `m_{water}=750 g`

We have that `Q=mc\Delta T`, so

`Q_{object}=-Q_{water}`

`m_{object} \quad c_{object} \quad \Delta T_{object}=-m_{water}\quad c_{water} \quad \Delta T_{water}`

Substitute known values and obtain: `\quad c_{object}=1.88J/{g\quad^o C}`

If you want the units of the answer to be in terms of calories, `\text{cal}/{g\quad^o C}`, use the unit conversion `1 cal=4.18J`

`c_{object}=0.45\text{cal}/{g\quad^o C}`