# What is the heat capacity #C_"cal"# of the dry calorimeter device?

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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

#### 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