D. Energy Changes and Rates of Reaction · Energy Changes and Rates of ReactionCombustion Calorimetry, Fuel Energy and Experimental Error
D2.3 D2.4 D3.3 A1.1 A1.2 A1.4 A1.8 A1.10 A1.13About 75 minutes
Learning goals- I can calculate a molar enthalpy of combustion from simple calorimetry data, including the heat absorbed by the container.
- I can calibrate a bomb calorimeter and use q = CΔT to find the energy content of a fuel or food.
- I can plan a calorimetry inquiry with a hypothesis, variables and safety precautions.
- I can compare an experimental value with an accepted value and explain specific sources of error.
Words to know
heat of combustion
The energy released when a substance burns completely in oxygen; per mole, the molar enthalpy of combustion ΔH_comb.
bomb calorimeter
A sealed steel container in which a sample burns in pure oxygen, surrounded by water in an insulated jacket; used for accurate heats of combustion.
calibration
Finding a calorimeter's total heat capacity by burning a substance with an exactly known energy, such as benzoic acid.
food Calorie (Cal)
The energy unit on food labels: 1 Cal = 1 kcal = 4.184 kJ.
controlled variable
A variable kept the same in all trials so that only the independent variable affects the result.
source of error
A specific reason a measured value differs from the true value, such as heat lost to the air.
Where do food Calories come from?
The Nutrition Facts table on a granola bar says something like "Calories 190". Someone, somewhere, burned food in a calorimeter to get numbers like that. Food chemists, fuel engineers and car makers all need reliable heats of combustion. In this lesson you will learn two ways to measure them: a simple can calorimeter you could build in a school lab, and the precise bomb calorimeter used by professionals. You will also learn to plan a calorimetry inquiry and to explain, honestly and specifically, why an experimental value differs from the accepted one.
The simple (can) calorimeter
A small burner of liquid fuel (or a candle) heats a metal can of water. You measure:
- the mass of water and its temperature change,
- the mass of the burner before and after (so you know the mass of fuel burned),
- optionally, the mass of the can (so you can include the heat it absorbs).
Then q_absorbed = (mcΔT)_water + (mcΔT)_can, q_fuel = −q_absorbed, and ΔH_comb = q_fuel ÷ n_fuel.
This method is simple but inaccurate: much of the heat misses the can. Typical school results are 20% to 50% smaller than the accepted values.
Worked exampleExample 1: Molar enthalpy of combustion of ethanol
A spirit burner's mass falls from 152.364 g to 151.482 g while it heats 200.0 g of water in a 15.0 g aluminum can from 19.0 °C to 41.5 °C. Find the molar enthalpy of combustion of ethanol, C₂H₅OH (M = 46.08 g/mol), and the percentage error (accepted value −1367 kJ/mol).
- ΔT = 41.5 − 19.0 = 22.5 °C.
- q_water = 200.0 × 4.18 × 22.5 = 18 810 J. q_can = 15.0 × 0.897 × 22.5 = 303 J. Total absorbed = 19 113 J = 19.11 kJ.
- Mass burned = 152.364 − 151.482 = 0.882 g. n = 0.882 ÷ 46.08 = 0.01914 mol.
- ΔH_comb = −19.11 kJ ÷ 0.01914 mol = −998 kJ/mol ≈ −999 kJ/mol (using unrounded values).
- % error = (−999 − (−1367)) ÷ (−1367) × 100% = −26.9%.
ΔH_comb ≈ −999 kJ/mol, about 27% smaller in size than the accepted −1367 kJ/mol. Most of the difference is heat that never reached the can.
The bomb calorimeter
For accurate values, the sample is burned inside a sealed steel container (the bomb) filled with pure oxygen under pressure. The bomb sits in a measured mass of water inside an insulated jacket. An electric current ignites the sample, and it burns completely and quickly. Almost no heat escapes.
Because the bomb, water, stirrer and thermometer all absorb heat, the whole calorimeter is calibrated first by burning a substance whose energy is known exactly (usually benzoic acid, which releases 26.43 kJ per gram). This gives the calorimeter's total heat capacity C in kJ/°C. Then:
q = CΔT
(Strictly, a sealed bomb measures the energy change at constant volume, which differs slightly from ΔH for reactions that change the amount of gas. The difference is usually small, and you will learn to correct for it in first-year university chemistry.)
Worked exampleExample 2: Calibrating a bomb calorimeter
Burning 1.000 g of benzoic acid (26.43 kJ/g) raises the temperature of a bomb calorimeter by 2.584 °C. Find the heat capacity of the calorimeter.
- Heat released = 1.000 g × 26.43 kJ/g = 26.43 kJ. All of it is absorbed by the calorimeter.
- C = q ÷ ΔT = 26.43 kJ ÷ 2.584 °C = 10.23 kJ/°C.
Worked exampleExample 3: Energy content of sugar
In the same calorimeter, 0.7500 g of sucrose, C₁₂H₂₂O₁₁ (M = 342.34 g/mol), raises the temperature by 1.209 °C. Find the molar enthalpy of combustion of sucrose and its energy per gram in kJ/g and in food Calories (1 Cal = 4.184 kJ).
- q = CΔT = 10.23 kJ/°C × 1.209 °C = 12.37 kJ absorbed, so 12.37 kJ released.
- n = 0.7500 g ÷ 342.34 g/mol = 0.002191 mol.
- ΔH_comb = −12.37 kJ ÷ 0.002191 mol = −5645 kJ/mol.
- Per gram: 12.37 kJ ÷ 0.7500 g = 16.49 kJ/g. In Calories: 16.49 ÷ 4.184 = 3.94 Cal/g.
ΔH_comb = −5645 kJ/mol, about 16.5 kJ/g, or 3.9 Cal per gram. That matches the familiar rule of thumb of about 4 Calories per gram of carbohydrate.
Planning a calorimetry inquiry
Suppose your question is: How does the molar enthalpy of combustion change from methanol to ethanol to 1-propanol? A good plan includes:
- Hypothesis with a reason: each extra CH₂ group adds one more C–C bond and two more C–H bonds to burn, so the energy released should increase by roughly the same amount each time.
- Variables: independent variable: the alcohol; dependent variable: ΔH_comb (from ΔT and mass of fuel burned); controlled variables: mass of water, can, distance from wick to can, starting water temperature, approximate ΔT, draft shielding.
- Safety (WHMIS): alcohols carry the flame pictogram (flammable). Keep stock bottles capped and away from flames, wear goggles, tie back hair, and never refill a burner that is lit or hot.
- Data table with units and the precision of each instrument.
- Analysis: calculate ΔH, compare with accepted values (methanol −726 kJ/mol, ethanol −1367 kJ/mol, 1-propanol −2021 kJ/mol), calculate percentage error, and identify specific sources of error with their effect (too high or too low).
The accepted values do increase by about 640 to 655 kJ/mol per CH₂ group, which supports the hypothesis. You will name these alcohols properly in the Organic Chemistry unit.
| Source of error | Effect on the measured size of ΔH | Improvement |
|---|
| Heat lost to the air around the can | too small | draft shield, lid, flame closer to the can |
| Heat absorbed by the can and thermometer | too small (unless included) | include q_can; use a calibrated calorimeter |
| Incomplete combustion (soot, CO) | too small | adjust the wick or air supply; bomb calorimeter with pure O₂ |
| Water produced as vapour, not liquid | too small compared with a value for H₂O(l) | compare with the value for H₂O(g), or condense the products |
| Fuel evaporating from the wick before weighing | too small (evaporated fuel is counted as burned, so n is too large) | cap the burner and weigh immediately |
Hands-on activityHow efficient is a microwave oven?
You need: a microwave oven with its output power printed on the label or in the manual (for example "1000 W"), a microwave-safe mug, a measuring cup, a kitchen thermometer, a spoon, oven mitts and a grown-up. Safety: the grown-up removes the hot mug with oven mitts; do not heat the water past about 70 °C; stir before measuring (microwaves heat unevenly).
- Measure 250 mL (250 g) of cold tap water into the mug and record its temperature.
- Heat at full power for exactly 60 s. Stir and record the temperature.
- Energy delivered by the oven = power × time (for example 1000 W × 60 s = 60 000 J).
- Energy absorbed by the water = mcΔT. Calculate the efficiency = (energy absorbed ÷ energy delivered) × 100%.
- List at least three specific reasons why the efficiency is below 100%, and say which way each one pushes your result.
What to look for: a typical result is somewhere around 60% to 85%. Energy also warms the mug, escapes from the water surface, and some of the rated power is never absorbed. If your thermometer reads the top of the water without stirring, your ΔT may be too high. This is the same kind of error analysis you write for a lab report.
Just for funA nutrition-label Calorie (capital C) is really a kilocalorie: 1 Cal = 1000 cal = 4.184 kJ. So a 190-Calorie granola bar holds about 795 kJ, enough energy, in principle, to warm a large pot of about 2 kg of water from fridge-cold to steaming hot. Please eat the granola bar instead.
Your turn
1Burning 0.500 g of a candle heats 150.0 g of water by 15.0 °C. Ignoring the can, find the heat released per gram of wax.
q = 150.0 × 4.18 × 15.0 = 9405 J = 9.41 kJ. Per gram: 9.41 ÷ 0.500 = 18.8 kJ/g released.
2A burner loses 1.15 g of methanol (M = 32.05 g/mol) while heating 250.0 g of water from 18.0 °C to 35.5 °C. Find ΔH_comb of methanol (ignore the can).
q = 250.0 × 4.18 × 17.5 = 18 288 J = 18.29 kJ. n = 1.15 ÷ 32.05 = 0.03588 mol. ΔH = −18.29 ÷ 0.03588 = −510 kJ/mol.
3The accepted value for methanol is −726 kJ/mol. Calculate the percentage error for the previous question.
(−510 − (−726)) ÷ (−726) × 100% = 216 ÷ −726 × 100% = −29.8% (the measured value is about 30% too small in size).
4A bomb calorimeter has C = 8.50 kJ/°C. Burning 0.400 g of a food raises the temperature by 1.60 °C. Find the energy per gram.
q = 8.50 × 1.60 = 13.6 kJ; per gram: 13.6 ÷ 0.400 = 34.0 kJ/g (a fatty food).
5Why must a bomb calorimeter be calibrated before use?
The steel bomb, water, stirrer and thermometer all absorb heat. Calibration with a known substance gives the total heat capacity C of the whole calorimeter, so q = CΔT.
6State two specific sources of error in a can calorimeter and the effect of each.
Any two from the table, for example: heat lost to the air (ΔH too small); incomplete combustion making soot (ΔH too small); heat absorbed by the can if not included (ΔH too small).
More practice: warm-up
1What does a bomb calorimeter burn the sample in?
Pure oxygen under pressure, inside a sealed steel container.
2What is the equation for heat in a calibrated calorimeter?
3How many kilojoules are in one food Calorie?
4Write the formula for percentage error.
(experimental − accepted) ÷ accepted × 100%
5What WHMIS hazard do alcohol fuels have?
They are flammable (flame pictogram).
More practice: on your own
1A calorimeter is calibrated by burning 0.800 g of benzoic acid (26.43 kJ/g); ΔT = 2.150 °C. Find C.
q = 0.800 × 26.43 = 21.14 kJ. C = 21.14 ÷ 2.150 = 9.83 kJ/°C.
2In that calorimeter (C = 9.83 kJ/°C), 0.250 g of octane, C₈H₁₈ (M = 114.26 g/mol), raises the temperature by 1.225 °C. Find ΔH_comb of octane.
q = 9.83 × 1.225 = 12.04 kJ. n = 0.250 ÷ 114.26 = 0.002188 mol. ΔH = −12.04 ÷ 0.002188 = −5.50 × 10³ kJ/mol (accepted about −5470 kJ/mol).
3A student's ethanol result is −1450 kJ/mol, larger in size than the accepted −1367 kJ/mol. Suggest one error that could make the value too large.
Any error that makes ΔT too large or the mass of fuel burned too small. For example: the thermometer touched the hot bottom of the can, so the final temperature read too high (ΔT and q too large); or the burner's final mass was recorded too high, so less fuel seemed to burn (n too small). Either makes ΔH too large in size.
4Why do simple calorimetry results for combustion usually have much larger percentage errors than coffee-cup results for neutralization?
In a coffee-cup calorimeter the reaction happens inside the water, so almost all the heat goes into it. A flame heats the air, the can and the room; much of the heat never reaches the water.
5A snack bar contains 9.0 g of fat (about 37 kJ/g), 25 g of carbohydrate (about 17 kJ/g) and 4.0 g of protein (about 17 kJ/g). Estimate its energy in kJ and in Calories.
Fat 9.0 × 37 = 333 kJ; carbohydrate 25 × 17 = 425 kJ; protein 4.0 × 17 = 68 kJ. Total ≈ 826 kJ, or 826 ÷ 4.184 ≈ 197 Cal (about 200 Calories).
Challenge
1A can calorimeter containing 200.0 g of water (with a 50.0 g aluminum can) is used with an ethanol burner. If the true ΔH_comb is −1367 kJ/mol and only 35.0% of the heat reaches the can and water, what temperature change will burning 1.00 g of ethanol produce?
n = 1.00 ÷ 46.08 = 0.02170 mol; heat released = 0.02170 × 1367 = 29.67 kJ; heat reaching water and can = 0.350 × 29.67 = 10.38 kJ = 10 383 J. Heat capacity of water + can = 200.0 × 4.18 + 50.0 × 0.897 = 836 + 44.85 = 880.9 J/°C. ΔT = 10 383 ÷ 880.9 = 11.8 °C.
2Design a controlled experiment to test whether a draft shield (a cylinder of aluminum foil around the flame and can) reduces heat loss. State the independent, dependent and controlled variables and how you would know the shield helped.
Independent: shield or no shield. Dependent: calculated ΔH_comb (or ΔT per gram of fuel). Controlled: same fuel, mass of water, can, distance from flame to can, starting temperature, burning time or target ΔT, room drafts. Repeat each condition at least three times. The shield helped if its average ΔH is larger in size and closer to the accepted value, by more than the spread between trials.
3The accepted ΔH_comb of ethanol is −1367 kJ/mol when the water formed is liquid, and −1235 kJ/mol when it is gas. In a can calorimeter, the water formed escapes as vapour. Which accepted value should students compare their result with, and why does it matter for the percentage error in Example 1?
Compare with −1235 kJ/mol, since the water leaves as vapour and its heat of condensation is never released into the can. Then the error in Example 1 is (−999 − (−1235)) ÷ (−1235) × 100% = −19.1% instead of −26.9%: part of the apparent "error" was really the wrong comparison.
Quiz
1. Why is a bomb calorimeter more accurate than a can calorimeter?
2. A calorimeter has C = 10.0 kJ/°C. Burning a sample raises its temperature by 2.50 °C. How much heat did the sample release?
3. A student's measured ΔH_comb is −850 kJ/mol; the accepted value is −1000 kJ/mol. What is the percentage error?
4. Which is a specific source of error that makes a can-calorimeter ΔH_comb too small in size?
5. Burning 0.500 g of a food releases 8.37 kJ. About how many food Calories is that per gram?
6. In a combustion inquiry comparing three alcohols, which should be a controlled variable?
Big ideas (D2.3, D2.4): the heat of combustion of a fuel is measured from the temperature rise of water (q = mcΔT) or of a calibrated calorimeter (q = CΔT), divided by the moles of fuel burned. Simple calorimeters lose much of the heat; bomb calorimeters are accurate. Good lab reports compare results with accepted values, calculate percentage error and name specific sources of error and their direction.
How to help: Do the microwave activity together and ask your teen to write three specific error sources. Read a Nutrition Facts table together and convert the Calories to kilojoules (×4.184).
Common difficulties:
- Using the mass of water instead of the mass of fuel burned to find moles.
- Forgetting the negative sign on ΔH_comb.
- Writing vague errors ("human error", "miscalculation"). Ask: what exactly went wrong, and did it make the answer too big or too small?
- Confusing calibration (finding C) with measurement (using C).
Activity safety: a grown-up handles the hot mug with oven mitts. Do not overheat water in a microwave (superheated water can boil suddenly); keep it below about 70 °C and stir. In school labs, alcohol burners are used only with goggles and teacher supervision; do not try burning fuels at home.