A 5.1 g piece of gold jewelry is removed from water at 100 degrees celcius and placed in a coffee-cup calorimeter containing 16.9 g of water at 22.5 degrees celcius. The equilibrium temperature of the water and jewelry is 23.2 degrees celcius. The calorimeter constant is known from calibration experiments to be 1.54 J/degrees celcius (without water). What is the specific heat of the piece of jewelry? Is the jewelry pure gold?
By principle of calorimeter,Heat loss is equals to Heat gain.
Gold looses heat and water in coffee cup calorimeter gains heat.
Mass of water cup = 16.9g
Initial temp. of coffee cup water= 22.5 deg. C
Specific heat of water= 4.184 J/g C
Mass of gold =5.1 g
Temp of gold =100 deg C
specific heat of gold = s (say) ?
Temp of equilibrium= 23.2 deg C
Calorimeter constant= 1.54 J/C
Heat given by gold =5.1 x s x (100-23.2)
=391.68 s J
Heat taken by water in coffee cup =16.9 x 4.184 x(23.2-22.5)
Heat taken by Calorimeter= 1.54 x (23.2-22.5)
Heat lost = Heat gain
391.68 s =49.496+1.078
s=.1291 J/g C
Yes it pure gold.
specific heat of gold= .129 J/(g C)