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A constant-pressure calorimeter contained 75 g of water at 16.95 degrees celcius. A...
A constant-pressure calorimeter contained 75 g of water at 16.95 degrees celcius. A 93.3 g sample of iron at 65.58 degrees celcius was placed in it, giving it a final temperature of 19.68 degrees celcius for the system. Calculate the heat capacity of the calorimeter (without water). Specific heats are 4.184 J/g degrees celcius for water and 0.444 J/g degrees celcius for Fe(s)
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By principle of calorimeter,Heat loss is equals to Heat gain.Iron looses heat and water in calorimeter gains heat.
mass of water= 75 g
initial temp. of water= 16.95 deg. C
specific heat of water= 4.184 J/g C
mass of iron= 93.3 g
temp of iron =65.58 deg C
specific heat of iron= .444 J/gC
temp of equilibrium= 19.68
Heat given by iron ball=93.3 x .444 x(65.58-19.68)
Heat taken by water =75 x 4.184 x(19.68-16.95)
Thus Heat taken by Calorimeter=1044.75 J
Heat capacity of calorimeter= (1044.75)/(2.73)
Posted by pramodpandey on May 7, 2013 at 10:34 AM (Answer #1)
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