AP Chemistry 6.3: Heat Capacity and Calorimetry

Different substances require different amounts of energy to change temperature.

• For example, water requires a relatively large amount of energy to increase its temperature compared with many metals. This is described using heat capacity and specific heat capacity.
• Khan Academy's current Unit 6 includes heat capacity, constant-pressure calorimetry, food-energy calorimetry, and constant-volume calorimetry.

The heat capacity, usually represented by C, is the amount of heat required to change the temperature of an entire object by 1°C or 1 K.

• Units might be:
• J/°C
• or:
• J/K
• A large heat capacity means the object requires a lot of energy for a small temperature change.

The specific heat capacity, often written c, is the amount of energy required to raise the temperature of 1 gram of a substance by 1°C.

• Typical units are:
• J/(g·°C)
• For liquid water:
• c ≈ 4.18 J/(g·°C)
• This means approximately 4.18 J are needed to increase the temperature of 1 g of liquid water by 1°C.

A molar heat capacity describes the heat required to increase the temperature of 1 mole of a substance by one degree.

• Its units may be:
• J/(mol·K)

Be careful to identify whether the problem provides specific heat per gram or heat capacity per mole.

The most important calorimetry equation is:

• q = mcΔT
• where:
• q = heat transferred
• m = mass
• c = specific heat capacity
• ΔT = temperature change
• and:
• ΔT = Tfinal − Tinitial

The sign of ΔT matters.

If temperature rises:

• ΔT > 0
• so q is positive for the substance gaining heat.

If temperature falls:

• ΔT < 0
• so q is negative for the substance losing heat.

Suppose 100.0 g of water warms from 20.0°C to 35.0°C.

• For water:
• c = 4.18 J/(g·°C)

Temperature change:

• ΔT = 35.0 − 20.0 = 15.0°C

Use:

• q = mcΔT
• q = (100.0 g)(4.18 J/(g·°C))(15.0°C)
• q = 6270 J
• or:
• 6.27 kJ
• The positive value makes sense because the water gained energy.

Suppose 50.0 g of a substance with:

• c = 0.900 J/(g·°C)
• cools from 80.0°C to 30.0°C.
• ΔT = 30.0 − 80.0 = −50.0°C
• Then:
• q = (50.0)(0.900)(−50.0)
• q = −2250 J
• The negative sign means the sample released heat.

Calorimetry is the experimental measurement of heat transfer.

A calorimeter is the device used to measure this energy transfer.

The fundamental assumption in an ideal calorimetry problem is conservation of energy:

• qreaction + qsurroundings = 0
• Therefore:
• qreaction = −qsurroundings
• This sign reversal is one of the most important things to remember.
• If the solution gains 5.00 kJ:
• qsolution = +5.00 kJ
• then the reaction released:
• qreaction = −5.00 kJ
• Therefore the reaction was exothermic.

A coffee-cup calorimeter is commonly used for reactions occurring in solution at approximately constant atmospheric pressure.

• Because the process occurs at constant pressure:
• qreaction,p = ΔHreaction

Usually the temperature change of the surrounding solution is measured.

• If we assume the solution behaves approximately like water:
• qsolution = msolution csolution ΔT
• Then:
• qreaction = −qsolution

If the calorimeter itself absorbs a significant amount of energy, its heat capacity may also need to be included:

• qreaction + qsolution + qcalorimeter = 0

Suppose a reaction occurs in 100.0 g of solution. The temperature rises from 21.0°C to 27.0°C. Assume:

• csolution = 4.18 J/(g·°C)

Temperature change:

• ΔT = 6.0°C

Heat gained by solution:

• qsolution = (100.0)(4.18)(6.0)
• qsolution = +2508 J
• The solution gained heat.

Therefore the reaction lost heat:

• qreaction = −2508 J
• or:
• −2.51 kJ
• The reaction is exothermic.

Calorimetry may give you heat for the amount that actually reacted, but AP questions often ask for:

• ΔH in kJ/molrxn
• Suppose the reaction released:
• −2.51 kJ
• and:
• 0.0250 mol
• of reaction occurred.
• Then:
• ΔHrxn = −2.51 kJ / 0.0250 mol
• ΔHrxn = −100.4 kJ/molrxn

Always use stoichiometry if necessary to determine how many moles of reaction occurred.

A bomb calorimeter is used for processes such as combustion reactions under constant volume.

• Instead of measuring ΔH directly, constant-volume calorimetry relates heat to the change in internal energy:
• qᵥ = ΔE

A bomb calorimeter often uses:

• qcal = CcalΔT
• where:
• Ccal = heat capacity of the calorimeter
• ΔT = calorimeter temperature change
• Then:
• qreaction = −qcal
• If the calorimeter gets warmer, it gained heat, meaning the combustion reaction released heat.

The equation:

• q = mcΔT
• can also be rearranged:
• c = q/(mΔT)
• Suppose a 50.0 g metal absorbs 450 J and warms by 20.0°C:
• c = 450 / [(50.0)(20.0)]
• c = 0.450 J/(g·°C)

Specific heat measurements can help identify an unknown material.

For an isolated system:

• qhot + qcold = 0
• If each uses q = mcΔT:
• mhot chot(Tf − Thot,i) + mcold ccold(Tf − Tcold,i) = 0
• You can solve this equation for the final equilibrium temperature.

Always use:

ΔT = final − initial

Do not switch the order depending on whether something heats or cools.

Remember that the reaction and surroundings have opposite q signs.

Do not automatically use only the mass of the solute in a coffee-cup experiment. The temperature change usually belongs to the entire solution.

Watch your units:

1000 J = 1 kJ

The main calorimetry chain is:

• temperature change → q = mcΔT → reverse sign → reaction heat → divide by moles if needed