AP® Chemistry review sheet from Aim for Five (aimforfive.com/chem/units/6/6-4)
Unit 6 · Topic 6.4
6.4 Heat Capacity and Calorimetry
The heat needed to change a substance's temperature depends on its mass, its specific heat capacity and the temperature change: q = mcΔT. Calorimetry uses this equation and the law of conservation of energy to measure the heat released or absorbed by a reaction or by dissolving.
Key terms
- specific heat capacity
- molar heat capacity
- calorimetry
- q = mcΔT
- first law of thermodynamics
Specific heat and molar heat capacity
Specific heat capacity, c, is the energy needed to raise the temperature of 1 g of a substance by 1 °C. Its units are J/(g·°C), which is the same as J/(g·K) because a change of 1 °C equals a change of 1 K.
Molar heat capacity is the same idea per mole instead of per gram, in J/(mol·°C). To convert, multiply the specific heat by the molar mass. For water, 4.18 J/(g·°C) × 18.02 g/mol ≈ 75.3 J/(mol·°C).
Substances with a high specific heat warm up slowly. Water's value is unusually high, which is why lakes and oceans change temperature slowly.
| Substance | Specific heat, J/(g·°C) |
|---|---|
| Liquid water | 4.18 |
| Aluminum | 0.897 |
| Iron | 0.449 |
| Copper | 0.385 |
Using q = mcΔT
q = mcΔT, where m is mass in grams, c is specific heat and ΔT = T(final) − T(initial). If the temperature rises, ΔT and q are positive (the substance gained energy). If it falls, both are negative.
With the same amount of energy, a substance with a smaller c gets a bigger temperature change. Adding 1000 J to 10.0 g of aluminum raises it by about 111 °C, but the same 1000 J raises 10.0 g of water by only about 23.9 °C.
The molar version works the same way: q = n × (molar heat capacity) × ΔT, with n in moles.
Conservation of energy in a calorimeter
The first law of thermodynamics says energy is conserved. A system can change its energy in three main ways: by being heated or cooled, by changing phase, or by reacting. In a calorimeter, all the energy a process releases is picked up by the surrounding solution, and all the energy a process absorbs comes out of it.
So q(process) = −q(solution). If the solution's temperature rises, q(solution) is positive and q(process) is negative: the process is exothermic. If the solution cools, the process is endothermic.
Coffee-cup calorimetry step by step
A simple calorimeter is a foam cup (a good insulator) with a lid and a thermometer, open to the air so the pressure is constant. That means the heat measured is the enthalpy change.
- Find q(solution) = mcΔT. Usually m is the total mass of the solution, and you assume c is the same as water's, 4.18 J/(g·°C), and the density is 1.00 g/mL unless told otherwise.
- Flip the sign: q(process) = −q(solution).
- Divide by the moles of the substance that reacted or dissolved to get ΔH in J/mol, then convert to kJ/mol.
- If the problem gives the calorimeter's own heat capacity (in J/°C), add C(cal) × ΔT to the heat absorbed by the solution.
Sources of error
Real foam cups leak a little energy. In an exothermic experiment, some heat escapes to the room, so the measured temperature rise is too small and the calculated ΔH is less negative than the true value. In an endothermic experiment, some heat leaks in from the room, so the temperature drop is too small and the calculated ΔH is less positive than the true value. Either way the magnitude of ΔH comes out too small.
Other common issues: not all of the solid dissolves, the thermometer is read before the temperature finishes changing, or some solution splashes out.
Worked examples
Try each one yourself first, then open the solution.
- Example 1Calculator allowed
Heat to warm water
How much heat is needed to warm 250. g of water from 20.0 °C to 80.0 °C? The specific heat of water is 4.18 J/(g·°C).
Show the solutionHide the solution
- Step 1: ΔT = 80.0 °C − 20.0 °C = 60.0 °C.
- Step 2: q = mcΔT = (250. g)(4.18 J/(g·°C))(60.0 °C) = 62,700 J.
- Step 3: Convert to kilojoules: 62,700 J ÷ 1000 = 62.7 kJ. The value is positive because the water gains energy.
Answer: q = 6.27 × 10⁴ J = 62.7 kJ
- Example 2Calculator allowed
Enthalpy of dissolving from calorimeter data
A student dissolves 4.00 g of NH₄NO₃ (molar mass 80.04 g/mol) in 60.0 g of water in a coffee-cup calorimeter. The temperature falls from 22.0 °C to 17.2 °C. Assuming the solution has a specific heat of 4.18 J/(g·°C) and no heat is exchanged with the room, calculate ΔH for dissolving NH₄NO₃ in kJ/mol.
Show the solutionHide the solution
- Step 1: Mass of solution = 60.0 g water + 4.00 g solid = 64.0 g. ΔT = 17.2 − 22.0 = −4.8 °C.
- Step 2: q(solution) = mcΔT = (64.0 g)(4.18 J/(g·°C))(−4.8 °C) = −1284 J. The solution lost energy.
- Step 3: The dissolving process absorbed that energy: q(dissolving) = −q(solution) = +1284 J.
- Step 4: Moles of NH₄NO₃ = 4.00 g ÷ 80.04 g/mol = 0.04998 mol.
- Step 5: ΔH = +1284 J ÷ 0.04998 mol = +25,700 J/mol = +25.7 kJ/mol. (ΔT has only two significant figures, so +26 kJ/mol is the strictly correct report.)
- Step 6: The positive sign matches the observation: the solution cooled, so dissolving is endothermic.
Answer: ΔH ≈ +26 kJ/mol (endothermic)
- Example 3Calculator allowed
Trap: using the right ΔT for each object
A 25.0 g sample of an unknown metal is heated to 100.0 °C and placed in 50.0 g of water at 20.0 °C. The final temperature of the water and metal is 23.4 °C. What is the specific heat of the metal? Use 4.18 J/(g·°C) for water.
Show the solutionHide the solution
- Step 1: The metal and water each have their own ΔT. Water: 23.4 − 20.0 = +3.4 °C. Metal: 23.4 − 100.0 = −76.6 °C. Using 3.4 °C for the metal is the trap.
- Step 2: q(water) = (50.0 g)(4.18)(3.4 °C) = 710.6 J.
- Step 3: The metal lost what the water gained: q(metal) = −710.6 J.
- Step 4: c(metal) = q ÷ (mΔT) = −710.6 J ÷ [(25.0 g)(−76.6 °C)] = 0.371 J/(g·°C).
- Step 5: The 3.4 °C change has only two significant figures, so report 0.37 J/(g·°C).
Answer: c ≈ 0.37 J/(g·°C)
Common mistakes
- Forgetting to flip the sign between the solution and the process. A temperature rise in the water means the reaction's q and ΔH are negative.
- Dividing by grams instead of moles when a problem asks for kJ/mol.
- Mixing joules and kilojoules. q = mcΔT gives joules when c is in J/(g·°C).
- Using the mass of the solute alone, or forgetting to include it, when the problem says to use the total mass of the solution. Read which mass the problem tells you to use.
On the exam
- Lab-based free-response questions often give calorimeter data and ask for ΔH, then ask how an error (heat loss, incomplete dissolving) would change the result. Say whether the calculated value would be too high or too low in magnitude and why.
- State your assumptions when you use them, such as the solution's specific heat being equal to water's.
Connected topics
Videos
Check yourself
4 questions on 6.4 Heat Capacity and Calorimetry. Pick an answer to see if you got it, and why.
How much heat is needed to warm 50.0 g of water from 22.0 °C to 30.0 °C? (The specific heat of water is 4.18 J/(g·°C).)
| Measurement | Value |
|---|---|
| Mass of metal | 50.0 g |
| Initial temperature of metal | 100.0 °C |
| Mass of water in the calorimeter | 100.0 g |
| Initial temperature of water | 20.0 °C |
| Final temperature of metal and water | 23.0 °C |
Hypothetical data. A student heats a metal sample in boiling water, then drops it into water in an insulated cup. The specific heat of water is 4.18 J/(g·°C).
How much heat does the water absorb?
What is the specific heat of the metal?
Suppose some heat escaped from the cup to the air during the experiment. How would this affect the calculated specific heat of the metal?
0 of 4 answered