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?

1 Answer
Jun 3, 2016

Answer:

#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.