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The amount of heat absorbed by the calorimeter in this reaction must be determined before you can calculate the enthalpy of the reaction that occurs inside of it.
To do this, suppose that you measure exactly 43.5 mL of cool water at 22.0°C and
place it into a calorimeter. Then you add 21.0 mL of warm water (initial temperature
of 37.0°C) to the calorimeter. The final temperature of the entire system is 25.5°C.
Compute the heat capacity of the calorimeter in J K–1 assuming that no heat leaks
into or out of the calorimeter.

Sagot :

pstnonsonjoku

The heat capacity of the calorimeter is 16.8 J/K.

What is the heat capacity of the calorimeter?

Now we know that the calorimeter is a device from which there is no heat loss or gain. In this case, the heat lost by the hot water is equal to the heat gained by the calorimeter.

Thus;

cwθ = ccθ

Where;

cc= heat capacity of the calorimeter

cw = heat capacity of water

θ == temperature change

Hence;

4.2  * (37 - 25) = cc (25 - 22)

cc = 4.2  * (37 - 25)/(25 - 22)

cc = 16.8 J/K

Learn more about heat capacity:https://brainly.com/question/1747943

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