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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?
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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)
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