# 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?

Jun 3, 2016

#### Answer:

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

#### Explanation:

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

$q = m \times s \times \Delta T$

where, $m = 18.69 g$ is the mass of the object,
$s$ is the specific heat capacity,
and $\Delta T = {T}_{f} - {T}_{i}$ 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.