# 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·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·g–1·˚C–1

A piece of iron (specific heat capacity = 0.452 J·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·g–1·˚C–1

##### 1 Answer

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#### Explanation

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**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

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