Calorimetry量热法

Big Idea · Energy (ENE)AP Science Practice 2 · Question and MethodAP link · Unit 6~40 min

Learning goals

  • Explain how a calorimeter works: the heat released or absorbed by the system equals the heat absorbed or released by the surroundings
  • Use calorimetry data to find the specific heat of a metal and the enthalpy change of a reaction
  • Describe how to measure the heat of neutralization and identify sources of error

Before you start

  • Specific heat and q = mcΔT (2.10.3)
  • Molarity (2.9.2)

How are food Calories measured?

The Calories on a nutrition label were first measured by burning food. A small dried sample is sealed in a steel container full of oxygen and ignited, and all the heat it releases goes into the water around the container. Weigh the water, measure how much its temperature rises, and q = mcΔT from 2.10.3 gives the heat. Measuring heat this way is called calorimetry量热法, and the instrument is a calorimeter量热计.

How a calorimeter works

A calorimeter is insulated, so almost no heat passes in or out. Whatever heat the process (the system) releases is absorbed by the water or solution inside (the surroundings):

qsystem=−qsurroundingsq_\text{system} = -q_\text{surroundings}

For a reaction in solution, the surroundings are mostly the solution itself:

qrxn=−qsoln=−msoln c ΔTq_\text{rxn} = -q_\text{soln} = -m_\text{soln}\, c\, \Delta T

If the solution warms up (ΔT > 0), the reaction released heat and qrxn < 0. If the solution cools, the reaction absorbed heat. Dividing by the moles that reacted gives the heat per mole of reaction, the ΔH of lesson 2.10.5.

The same idea measures the specific heat of a metal: drop a hot block of metal into water, and the heat the metal loses equals the heat the water gains.

In US school labs, a coffee-cup calorimeter简易量热计 is two nested foam cups with a lid, a thermometer, and a stirrer. It works at constant (atmospheric) pressure. Heats of combustion are measured in a bomb calorimeter弹式量热计, a sealed steel vessel in which the reaction happens at constant volume.

The calorimeter in Chinese textbooks has the same parts as a coffee cup, just made of glass: an inner cup, an outer case with insulation between them, a lid, a thermometer, and a glass stirrer.

Figure: a simple calorimeter
thermometerglass stirrerlidinner cupinsulationouter case

The insulation and the lid keep heat from escaping, and the ring-shaped glass stirrer moves up and down so the whole solution is at the same temperature. A coffee-cup calorimeter uses two nested foam cups instead and works the same way.

For a strong acid neutralized by a strong base in dilute solution, forming 1 mol of H₂O(l) releases about 56 kJ: for H⁺(aq) + OH⁻(aq) → H₂O(l), ΔH = −55.8 kJ/mol.

This lesson uses the atomic masses on the AP periodic table (H 1.008, N 14.01, O 16.00, Na 22.99, Cl 35.45).

Try three experiments: hot metal in water, neutralization, and dissolving a salt. Change the data and watch the thermometer, then see how the measured temperature leads back to a specific heat or a ΔH. In the neutralization, try poor insulation:

Try it: a coffee-cup calorimeter
Metal

    How Chinese textbooks differ

    • The Chinese textbook’s calorimeter has an inner cup, an outer case, insulation between them, a lid, a thermometer, and a glass stirrer. The standard Chinese lab mixes 50 mL of 0.50 mol/L HCl with 50 mL of 0.55 mol/L NaOH, with a slight excess of base so that all the acid reacts.
    • Chinese students average the starting temperatures of the acid and the base, record the highest temperature reached, and repeat the experiment two more times.
    • The Chinese textbook gives the heat of neutralization as −57.3 kJ/mol, a little larger in size than the −55.8 kJ/mol used here. Sources differ slightly; use whatever value a problem gives.

    Example 1: specific heat of a metal

    A 200.0 g block of metal at 100.0 °C is dropped into 100.0 g of water at 22.0 °C. The final temperature is 34.1 °C. Find the specific heat of the metal. What metal could it be?

    Step-by-step solution

    1. The principle. Ignoring heat loss, all the heat the metal releases goes into the water: qmetal + qwater = 0. Find the heat from the water, then the metal’s specific heat.

    2. Heat absorbed by the water. For the water, ΔT = 34.1 − 22.0 = 12.1 °C:

      qwater=100.0×4.18×12.1=5.06×103 J\begin{aligned} q_\text{water} &= 100.0 \times 4.18 \times 12.1 \\ &= 5.06 \times 10^{3}\ \text{J} \end{aligned}

    3. Heat released by the metal. Opposite sign: qmetal = −5.06 × 10³ J.

    4. Temperature change of the metal.

      ΔT=34.1−100.0=−65.9 °C\Delta T = 34.1 - 100.0 = -65.9\ \text{°C}

    5. Specific heat.

      c=qmΔT=−5.06×103200.0×(−65.9)=0.384 J/(g⋅°C)\begin{aligned} c &= \frac{q}{m\Delta T} \\ &= \frac{-5.06 \times 10^{3}}{200.0 \times (-65.9)} \\ &= \boxed{0.384\ \text{J/(g·°C)}} \end{aligned}

    6. Conclusion. That is very close to copper’s 0.385 J/(g·°C), so the metal is probably copper.

    7. Check the signs. A negative divided by a negative is positive. A specific heat is always positive; if you get a negative one, one of the ΔT values is backward.

    Example 2: heat of neutralization

    In a coffee-cup calorimeter, 50.0 mL of 1.00 M HCl and 50.0 mL of 1.00 M NaOH, both at 21.0 °C, are mixed. The highest temperature reached is 27.6 °C. Assume the solution has a density of 1.00 g/mL and a specific heat of 4.18 J/(g·°C). Find ΔH per mole of water formed.

    Step-by-step solution

    1. The idea. The heat released by the neutralization is absorbed by the solution. Find that heat from the solution’s temperature change, then divide by the moles of water formed.

    2. Mass of the solution. The total volume is 50.0 mL + 50.0 mL = 100.0 mL, and the density is 1.00 g/mL, so m = 100.0 mL × 1.00 g/mL = 100.0 g.

    3. Temperature change.

      ΔT=27.6−21.0=6.6 °C\Delta T = 27.6 - 21.0 = 6.6\ \text{°C}

    4. Heat absorbed by the solution.

      qsoln=mcΔT=100.0×4.18×6.6=2.76×103 J=2.76 kJ\begin{aligned} q_\text{soln} &= mc\Delta T \\ &= 100.0 \times 4.18 \times 6.6 \\ &= 2.76 \times 10^{3}\ \text{J} \\ &= 2.76\ \text{kJ} \end{aligned}

    5. Moles of water formed. The acid and base are in equal amounts:

      n(HCl)=1.00×0.0500=0.0500 moln(NaOH)=1.00×0.0500=0.0500 mol\begin{aligned} n(\ce{HCl}) &= 1.00 \times 0.0500 \\ &= 0.0500\ \text{mol} \\ n(\ce{NaOH}) &= 1.00 \times 0.0500 \\ &= 0.0500\ \text{mol} \end{aligned}

      so 0.0500 mol of H₂O forms.

    6. ΔH. The reaction releases what the solution absorbs, so qrxn = −2.76 kJ:

      ΔH=−2.76 kJ0.0500 mol=−55 kJ/mol\begin{aligned} \Delta H &= \frac{-2.76\ \text{kJ}}{0.0500\ \text{mol}} \\ &= \boxed{-55\ \text{kJ/mol}} \end{aligned}

      ΔT has only two significant figures, so the answer has two.

    7. Compare. The reference value is −55.8 kJ/mol. Heat lost to the surroundings and absorbed by the cup usually makes measured values slightly smaller in size.

    8. A common mistake. Using the mass of only one solution (50.0 g), or forgetting that ΔH is negative when the solution warms up.

    Practice: from the calorimetry result, is ΔH for the reaction or process negative or positive?
    1. The solution warms from 20 °C to 26 °C.
    2. The solution cools from 22 °C to 15 °C.
    3. You calculate q_soln = +2.5 kJ.
    4. You calculate q_soln = −1.2 kJ.
    5. The outside of the calorimeter feels cold.
    6. Magnesium is added to hydrochloric acid and the thermometer reading rises.
    7. NH₄NO₃ dissolves and the solution gets colder.
    8. A sample burns in a bomb calorimeter and the surrounding water warms up.

    Common mistakes

    Mistake 1: “The solution got warmer, so the reaction absorbed heat.” What warmed up is the solution, the surroundings. The reaction released heat into it: qrxn = −qsoln < 0.

    Mistake 2: “Use the mass of only one of the solutions.” All of the mixed solution absorbs the heat: 50.0 mL of acid plus 50.0 mL of base is about 100.0 g.

    Mistake 3: “A metal stirrer works just as well as a glass one.” Metal conducts heat away quickly, so more heat escapes and the measured heat release is too small. Leaving the lid off or mixing slowly has the same effect.

    Quick check

    A reaction runs in 50.0 g of aqueous solution, which warms from 22.0 °C to 28.0 °C. Using c = 4.18 J/(g·°C), how much heat does the reaction release?

    In a neutralization experiment, some heat escapes through the walls of the calorimeter. How does this affect the calculated value of |ΔH|?

    Summary

    • A calorimeter stops heat from leaving, so qsystem = −qsurroundings. For a reaction in solution, qrxn = −msolncΔT.
    • For hot metal dropped into water, the heat released by the metal equals the heat absorbed by the water, which gives the metal’s specific heat.
    • Dividing qrxn by moles reacted gives ΔH per mole. Heat lost to the surroundings makes the measured |ΔH| too small.
    On to AP (for later)

    AP topic 6.4 builds on calorimetry: use qsystem = −qsurroundings to find a specific heat, a ΔH, or the final temperature of a mixture. Lab questions often ask how heat absorbed by the calorimeter, heat lost to the room, or spilled solution would change the calculated ΔH. Say clearly what happens to qsoln and to |ΔH|.

    Data sources: ΔH = −55.8 kJ/mol for H⁺(aq) + OH⁻(aq) → H₂O(l) is calculated from this site’s table of enthalpies of formation (lesson 2.10.6, NBS data).

    Practice this chapter: 2.10 · Thermochemistry · 10 questions

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