calorimetry

Calorimetry is the study of energy transfers (heat energy - represented as q or H, unit: Joule) between a system and its surroundings. Usually a "system" is a mass of water, or a block of metal, or any material that is at the center of the problem or question of study. The "surroundings" are typically ignored and treated as the //universe//, which means the heat energy is dissipated. In //heat exchange problems//, energy is transferred from one system to another.

__Systems__

There are three main calorimetry equations used when studying systems:
 * heating/cooling of a material: **q = m x Cp x (Tf - Ti)** where m is mass in grams, Cp is specific heat capacity in J/g C, Tf is final temperature and Ti is initial temperature in C or K.
 * melting/freezing of a material: **q = m x Hfus** where Hfus is the //enthalpy of fusion// in J/g
 * evaporating/condensing: **q = m x Hvap** where Hvap is the //enthalpy of vaporization// in J/g

These equations can be visualized on a __Heating Curve__:



The following sections on the graph are important:
 * **A - solid:** heating/cooling the solid phase (note both change in T and H)
 * **B - solid/liquid:** represents the phase change between the solid and liquid, where there is no change in T
 * **C - liquid:** heating/cooling the liquid phase (again changes in T and H)
 * **D - liquid/gas:** represents the phase change between the liquid and gas, where there is no change in T
 * **E - gas:** heating/cooling the gas phase (changes in T, H)

Note the lack of a (Tf - Ti) in the equations that involve phase changes (both for Hfus and Hvap). Phase changes occur at a constant temperature - this is how we assign melting and boiling points.

__Example 1:__

Calculate the heat energy required to warm a 200. g glass of water from 20 C to 80 C.

The key to calorimetry problems is the //verb//. In this question, the verb is "warm", which means the water is being heated. There is also an initial and final temperature provided.

Use the equation **q = m x Cp x (Tf - Ti)**; where Cp for water is 4.184 J/g C:

q = (200 g) x (4.184 J/g C) x (80 - 20) q = 50,208 J of energy

__Example 2:__

How much energy is lost from a 200 g glass of water at 23 C when it is cooled to 0 C, then frozen completely?

The verbs are "cooled" and "frozen", which require two equations: **q1 = m x Cp x (Tf - Ti)** and **q2 = m x Hfus**

Work out the two equations separately:

q1 = (200 g) x (4.184 J/g C) x (0 - 23) q1 = -19,246 J

q2 = (200 g) x (-333 J/g) (use a negative Hfus or Hvap when cooling/freezing - note the graph above) q2 = -66,600 J

Finally, sum the individual heat energies: q1 + q2 = -19246 + -66,600 = -85,846 J lost in the process.

__Example 3:__

This will be a heat exchange problem. In these problems, both substances will end up at the same temperature when the system reaches equilibrium.

A 25.0 g block of iron at 400 C is cooled by immersing it in 1000 g of water, initially at 20 C. When the systems equilibrate, what will be the final temperature of the two materials?

When dealing with a heat exchange problem, set two heating equations equal, with a negative sign in front of the substance that is cooling:

q (water) = - q (iron) (1000 g) x (4.184 J/g C) x (Tf - 20) = - (25.0 g) x (0.449 J/g C) x (Tf - 400)
 * m x Cp x (Tf - Ti) = -****m x Cp x (Tf - Ti)**

There are a few ways of solving for Tf; here is one: 4184 x (Tf - 20) = - 11.23 x (Tf - 400) 372.7 (Tf - 20) = - (Tf - 400) 372.7 Tf - 7455 = -Tf + 400 373.7 Tf = 7855 Tf = 21.0 C

The water warmed 1 C. This is not unusual - water has a very high heat capacity in relation to iron, and there was much more water than iron.

Try some practice problems: