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Why does the heat capacity of a calorimeter, C(cal), need to be determined?

A calorimeter contains 50.0 mL of water; both are at 21.2 degrees C. Then the 50.0 mL of water at 57.6 degrees C is added to the calorimeter. At equilibrium, the final temperature of the water and the calorimeter is 33.3 degrees C. Calculate the heat capacity, C, of the calorimeter.

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  • $\begingroup$ What is the mathematical equation used to determine the specific heat of a material used in calorimetry experiments? $\endgroup$ – bobthechemist Oct 22 '13 at 0:48
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The goal of calorimetry is to measure heat, that means flow of energy. This is impossible to do directly. But you can measure temperature, using thermometer.

So the basic question of calorimetry is: If I put 1 Joule, ($Q=1 J$) of energy into my object, how much the temperature changes? Of course, temperature $T$ increases and the proportionality constant is called Heat capacity $C_v$, then $Q = C_v \cdot \Delta T$

The net energy in closed system is zero, so heat lost equals heat gained.

And to your question - such experiment must be carried out in well isolated vessel. This is the calorimeter. It comes to thermal equilibrium as well and therefore its heat capacity must be known as well. This is usually established by experiment you described, so that everything is known and you can easily obtain the only unknown - $C_{calorimeter}$

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