TEKS
6G: Analyze and explain everyday examples that illustrate the laws of thermodynamics, including the law of conservation of energy and the law of entropy.
Objective
 Students will understand how calorimeters are used to find the specific heat of a substance by the heat transfer of a substance and how the units of calories are used to measure the energy of the heat transfer.
Equations
m1c1\begin{align*}\triangle\end{align*}t=m2c2\begin{align*}\triangle\end{align*}t
Though not particularly beautiful machines, calorimeters are incredibly useful ones. They are used to determine the calories (food energy) in food, as well as the average heat yield from burning various grades of coal and oil. The price of coal is often dependent on the heat yield from samples burned in a calorimeter.
Calorimetry
A calorimeter is a device used to measure changes in thermal energy or heat transfer. More specifically, it measures calories. A calorie is the amount of energy required to raise one gram of water by one degree Celsius. As such, the calorimeter measures the temperature change of a known amount of water. If a reaction is carried out in the reaction vessel, or if a measured mass of heated substance is placed in the water of the calorimeter, the change in the water temperature allows us to calculate the change in thermal energy.
The function of the calorimeter depends on the conservation of energy in a closed, isolated system. Calorimeters are carefully insulated so that heat transfer in or out is negligible. Consider the following example.
Example Problem: A 0.500 kg sample of water in a calorimeter is at 15.0ºC. A 0.0400 kg block of zinc at 115.0ºC is placed in the water. The specific heat of zinc is 388 J/kg•ºC. Find the final temperature of the system.
Solution: The heat lost by the block of zinc will equal the heat gain by the water in the calorimeter. In order to set heat gain mathematically equal to heat loss, either one of the terms must be made negative or the temperature change must be reversed. You should also note that the final temperature of the water and the block of zinc will be the same when equilibrium is reached. Using the equation for specific heat Q=mc\begin{align*}\triangle\end{align*}T and setting the heat lost by the zinc equal to the heat gained by the calorimetry container you can solve for the unknown variable.
\begin{align*}m_Wc_W(t_2  t_1)_W = m_{Zn}c_{Zn}(t_1  t_2)_{Zn}\end{align*}
\begin{align*}(0.500 \ \text{kg})(4180 \ \text{J/kg}^\circ \text{C})(x  15.0^\circ \text{C}) = (0.0400 \ \text{kg})(388 \ \text{J/kg}^\circ \text{C})(115.0^\circ \text{C}  x)\end{align*}
\begin{align*}2090 \ x  31350 = 1785  15.52 \ x\end{align*}
\begin{align*}2105.52 \ x = 33135\end{align*}
\begin{align*}x = 15.7^\circ \text{C}\end{align*}
Example Problem: A 100. g block of aluminum at 100.0ºC is placed in 100. g of water at 10.0ºC. The final temperature of the mixture is 25.0ºC. What is the specific heat of the aluminum as determined by the experiment?
Solution:
\begin{align*}m_Wc_W(t_2  t_1)_W = m_{Al}c_{Al}(t_1  t_2)_{Al}\end{align*}
\begin{align*}(0.100 \ kg)(4180 \ J/kg^\circ \text{C})(25.0^\circ \text{C}  10.0^\circ \text{C}) = (0.100 \ \text{kg})(x)(100.0^\circ \text{C}  25.0^\circ \text{C})\end{align*}
\begin{align*}6270 = 7.50 \ x\end{align*}
\begin{align*}x = 836 \ \text{J/kg}^\circ \text{C}\end{align*}
Summary

A calorimeter is a device used to measure changes in thermal energy or heat transfer.

If a reaction is carried out in the reaction vessel or if a measured mass of heated substance is placed in the water of the calorimeter, the change in the water temperature allows us to calculate the change in thermal energy.
Vocabulary
calorie: the amount of energy required to raise one gram of water by one degree Celsius.
calorimeter: a device used to measure changes in thermal energy or heat transfer.
Review
Questions

A 300.0 g sample of water at 80.0ºC is mixed with 300.0 g of water at 10.0ºC. Assuming no heat loss to the surroundings, what is the final temperature of the mixture?

A 400.0 g sample of methanol at 16.0ºC is mixed with 400.0 g of water at 85.0ºC. Assuming no heat loss to the surroundings, what is the final temperature of the mixture? The specific heat of methanol is 2450 J/kg•ºC.

A 100.0 g brass block at 100.0ºC is placed in 200.0 g of water at 20.0ºC. The specific heat of brass is 376 J/kg•ºC. Assuming no heat loss to the surroundings, what is the final temperature of the mixture?
After reviewing the questions above, click this link to practice a few more questions on calorimetry.
Here is a real world application on measuring the calories of everything you eat.
http://www.ck12.org/physics/CalorimetryinPhysics/rwa/MeasuringTheCaloriesofEverythingYouEat/
Practice Problems
Questions
The following video covers the calorimetry equation. Use this resource to answer the questions that follow.
 What is the number 4.18 J/g°C in the video?
 In the equation, q = mcΔt, what does c represent?
 What does it mean if the temperature in the calorimeter goes down?
Solved calorimetry problems:
http://calorimetryphysicsproblems.blogspot.com/2010/10/specificheatproblems.html
Laboratory:
The following on line simulation lab is an excellent opportunity to keep your hands dry and experience the techniques used to calculate the specific heat of metals using calorimeter:
http://www.chm.davidson.edu/vce/calorimetry/specificheatcapacityofcopper.html