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)

=1901.42 J

Heat taken by water =75 x 4.184 x(19.68-16.95)

=856.67 J

Thus Heat taken by Calorimeter=1044.75 J

**Heat capacity of calorimeter= (1044.75)/(2.73)**

**=382.69 J/C**

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