What is the heat capacity #C_"cal"# of the dry calorimeter device?
A piece of iron (specific heat capacity = #"0.452 J"cdot"g"^(-1)""^@ "C"^(-1)# ) with a mass of #"54.3218 g"# is heated in a water bath until it reaches a temperature of #98.2^@ "C"# . The iron is then transferred into a calorimeter (nested Styrofoam cups) that contains #"68.5314 g"# of water initially at #22.1^@ "C"# #.
The final temperature of the entire system is #27.5^@ "C"# . The specific heat capacity of water is #"4.184 J"cdot"g"^(-1)""^@ "C"^(-1)#
A piece of iron (specific heat capacity =
The final temperature of the entire system is
1 Answer
Warning! Long Answer.
Explanation:
We must identify the heat transfers that are occurring here.
One is the heat lost by the iron as it cools down (
A second is the heat gained by the water as it warms up (
A third is the heat gained by the calorimeter as it warms up (
Thus, the heat lost by the iron goes into warming up the water and the calorimeter.
One formula for the heat
#color(blue)(bar(ul(|color(white)(a/a)q = mCΔTcolor(white)(a/a)|)))" "#
where
A useful formula for the calorimeter is
#color(blue)(bar(ul(|color(white)(a/a)q = C_text(Cal)ΔTcolor(white)(a/a)|)))" "#
where
Per the Law of Conservation of Energy, the sum of the three heat transfers must be zero.
In this problem
Note: The answer can have only two significant figures because that is all we have for