Calculating Energy Changes

Chemical systems undergo three main processes that change their energy: heating/cooling, phase transitions, and chemical reactions. The way we calculate energy depends upon which of these processes we are performing. Because the change in energy associated with a given process is proportional to the amount of substance undergoing that process, this change is best described on a per mole (or per gram) basis.

Calorimetry is an experimental technique that is used to measure the change in energy of a chemical system. It is used experimentally to determine heat capacities or enthalpies of physical and chemical transformations. When you finish this module, you should be able to use calculations or estimations to relate energy changes associated with heating/cooling a substance to the heat capacity, relate energy changes associated with a phase transition to the enthalpy of fusion/ vaporization, and relate energy changes associated with a chemical reaction to the enthalpy of the reaction.

Calculating Energy Changes – Heating and Cooling

Heating a system increases the energy of the system, while cooling a system decreases the energy. When two systems are in thermal contact with each other and are otherwise isolated, the energy that comes out of one system is equal to the energy that goes into the other system. Eventually, thermal equilibrium is reached as the molecular collisions continue. This means that both systems will ultimately reach the same temperature. Energy changes that involve heating and cooling are governed by using the following equation that is based on the conservation of energy.


q lost = q gained
You will sometimes see this equation written with a negative sign, like this:
q lost = -q gained
If you calculate ΔT correctly, however, paying attention to the signs, this negative sign will already be present in your equation.

Heat Capacity

In addition to using q lost = q gained, there are several other equations we will use to calculate heat. One of these equations uses heat capacity. Heat capacity is the amount of heat required to change the temperature of a substance by one degree. Heat capacity depends upon the size of the sample (therefore an extensive property) and the identity of the substance.


q = CΔT

C is heat capacity (J/°C)


 A chemical system is put in thermal contact with a heat bath. The heat bath is a substance, such as water, whose heat capacity has been well established by previous experiments. A process is initiated in the chemical system (heating/cooling, phase transition, or chemical reaction), and the change in temperature of the heat bath is measured.

Because the heat capacity of the heat bath is known, the observed change in temperature can be used to determine the amount of energy exchanged between the system and the heat bath.

The energy exchanged between the system and the heat bath is equal in magnitude to the change in energy of the system. If the heat bath increased in temperature, its energy increased, and the energy of the system decreased by this amount. If the heat bath decreased in temperature, and therefore energy, the energy of the system increased by this amount.

Make sure to visualize this setup as you read and solve calorimetry problems. It will help you solve the problem correctly and understand what is happening conceptually.

Example #1

A piece of metal at 220°C is dropped into a cup containing 250 g of water. The water goes from 25.0 °C to 30.0 °C. What is the heat capacity of the piece of metal?

First, organize your data.


Metal Water
Ti = 220 °C Ti = 25.0 °C
Tf = 30.0 °C Tf = 30.0 °C
ΔT = 30 – 220 °C ΔT  = 30 – 25 = 5.0 °C
C = ? m = 250 g

Note:  The final temperature is the same for the metal and the water!

Next, think about your master equation.

q lost by metal = q gained by water

If you plug in the equation, q = CΔT, you get:

C(metal) (-190°C) = C(water)(5.0 °C)

In order to calculate C(metal), we need to know C(water). To do this, use the mass of the water and the definition of the calorie.  The calorie is the amount of energy required to raise 1 gram of water by 1 °Celsius.  Since we have 250 g of water, 250 cal is required for each degree.  We can make this into a conversion fact, just for this specific problem:

250 cal = 1 °C.

Heat capacity has the unit J/°C.  So we will also need to convert the energy to J, remembering the conversion fact: 1 cal = 4.184 J.

Now we can plug this in to the master equation above.

C(metal) (-190°C) = (1046 J)(5.0 °C)

C(metal) = 27.5 J/°C


A 12.48 g sample of an unknown metal, heated to 99.0 °C was then plunged into a 50.0 mL of 25.0 °C water. The temperature of the water rose to 28.1 °C.  Assuming no loss of energy to the surroundings:

  1.  How many joules of energy did the water absorb?
  2. How many joules of energy did the metal lose?
  3. What is the heat capacity of the metal?

1. q = (50.0 g)(3.1°C)(4.184 J g-1°C-1) = 648.52 J

2. 648.52 J

3. 648.52 J/ 70.9°C = 9.147 J/°C  .



Specific Heat Capacity

Specific heat capacity is an intensive property that is inherent to that substance, and does not change with the size of the sample. It is defined as the amount of energy required to raise the temperature of 1 gram of a substance 1 degree Celsius. It is important to understand that the heat capacity of a drop of water is much lower than the heat capacity of a pool of water, while their specific heats are the same. Note in the table here of some specific heats that the specific heat of water is much larger than other substances. You will be expected to memorize the specific heat of water. The value typically used is 4.184 J/g°C.


q = msΔT

s is specific heat capacity (J/g°C)

Sometimes lowercase c is used for specific heat capacity instead of s.  Just pay attention to the units to tell the difference between c and C.


Watch this video from ChemTeam that shows you how to calculate the specific heat of a metal. Make sure to solve the problem along with the video. You will see problems like this again!


For more examples of calculations involving specific heat, CLICK HERE to watch these other videos from the Chem Team. The videos are available on the right side of the Thermochemistry page.

    1. Calculate the final temperature when two water samples are mixed I.
    2. Calculate the final temperature when two water samples are mixed II.
    3. determine the specific heat of lead.
    4. How to calculate a calorimeter constant I.
    5. How to calculate a calorimeter constant II.


Suppose a piece of iron with a mass of 21.5 g at 100.0°C is dropped into an insulated container of water. The mass of the water is 132.0 g and its temperature before adding the iron is 20.0°C. What will be the final temperature of the system? Use the chart above for the specific heat of iron.

Click to reveal the answers.

qlost, metal = qgained, water
(mass)(Δt) (Cp, metal) = (mass)(Δt) (Cp, water)
(21.5g)(100°C – x°C)(0.45J/g°C) = (132.0g)(x°C – 20°C)(4.184J/g°C)
(2150 – 21.5x)(0.45) = (132x – 2640)(4.184)
965.5 – 9.675x = 552.288x – 11045.76
561.958x = 12011.26
To 3 sig figs, the answer is 21.4 °C.