Question

If it takes 41.72 joules to heat a piece of gold weighing 18.69 g from 10.0°C to 27.0°C, what is the specific heat of the gold?

Answer 1

`s=0.131J/(g*""^@C)`

Explanation:

The amount of heat (`q`) absorbed by gold could be expressed as:

`q=mxxsxxDeltaT`

where, `m=18.69g` is the mass of the object,
`s` is the specific heat capacity,
and `DeltaT=T_f-T_i` is the change on temperature of the sample.

Thus, the specific heat capacity is found by:

`s=q/(mxxDeltaT)=(41.72J)/(18.69gxx(27.0^@C-10.0^@C))=0.131J/(g*""^@C)`

Thermochemistry | Enthalpy and Calorimetry.