Module 4 · L2 of 13~35 min⚡ +50 XP in Learn · +25 to complete
Calorimetry, Combustion
In 1780, Antoine Lavoisier and Pierre-Simon Laplace used an ice calorimeter at the French Academy of Sciences to measure the heat of combustion of charcoal at approximately 34 kJ g⁻¹, one of the first quantitative measurements of chemical energy ever made. Their device surrounded the burning sample with ice and measured the mass of meltwater, turning the spirit of q = mcΔT into a physical scale. Every school calorimetry experiment today follows the same logic they established 245 years ago.
Today's hook, In 1780, Lavoisier and Laplace used an ice calorimeter at the French Academy of Sciences to measure charcoal's heat of combustion at ~34 kJ g⁻¹, the first quantitative measurement of chemical energy. The same principle, burn fuel, heat water, watch the thermometer, is what you'll use today.
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Warm up and recall
Warm up first
Three quick questions from earlier lessons. Pulling old material back to mind before you learn something new makes the new material stick better, so this is not busywork.
Worksheets
Practise this lesson
Four printable worksheets that build from the foundations up to exam-style questions, start at whatever level suits you.
Lesson question: how do you turn a thermometer reading into a single number, the molar enthalpy of combustion, that says exactly how much energy a fuel releases when it burns?
1Run the calorimeter. Burn a weighed fuel under a known mass of water and record the temperature rise.
2Find the heat. Use q = mcΔT, where m is the mass of the water, to get the heat the water gained.
3Scale to one mole. Divide by moles of fuel, ΔHc = −q/n, then compare with the accepted value and explain the gap.
Quick prerequisite: from L01, a negative ΔH means an exothermic reaction that releases heat, and combustion is always exothermic. If that is shaky, start with the Supported route later in this lesson. Syllabus reference: investigate enthalpy changes using calorimetry and q = mcΔT, comparing experimental results with reliable secondary data and explaining differences (ACSCH037, ACSCH073).
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Know what matters most
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Know what matters most
Must know
q = mcΔT, where m is the mass of water and c = 4.18 J g⁻¹ K⁻¹
ΔHc = −q/n; the sign is negative because combustion is exothermic
The named systematic errors and why each makes ΔHc less negative
The spirit burner components and the order of the procedure
Going deeper
Why a bomb calorimeter (constant volume) gives the accepted value
The ice-calorimeter method, heat from the mass of meltwater
Warm up and recall
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Recall, your gut answer first
+5 XP warm-up
Why does a small block of chocolate contain more kilojoules than a whole bowl of lettuce? The answer isn't just about quantity, it's about what the molecules are made of and how much energy is stored in their bonds.
If you burned both in a laboratory, you could measure exactly how much heat each released. How would you design that experiment? What would you actually be measuring, and what would you need to keep constant to make a fair comparison?
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What you'll master, and the words for it
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What you'll master
Know
Key Facts
The formulas q = mcΔT and ΔHc = −q/n
The components of a spirit burner calorimeter
Five sources of error in combustion calorimetry
Understand
Concepts
Why 'm' in q = mcΔT is the mass of water, not fuel
Why the negative sign appears in ΔHc = −q/n
Why experimental ΔHc is usually less negative than the accepted value
Can Do
Skills
Calculate q and then ΔHc from experimental data in 5 steps
Calculate percentage error between experimental and accepted values
Identify sources of error and state their directional effect on ΔHc
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Key terms
Calorimetry
The experimental measurement of heat changes in a chemical reaction using a calorimeter.
Heat of combustion (ΔHc)
The enthalpy change when one mole of a substance burns completely in oxygen under standard conditions.
q = mcΔT
The heat equation: heat (J) = mass of solution (g) × specific heat capacity (J g⁻¹ °C⁻¹) × temperature change.
Specific heat capacity (c)
Energy required to raise 1 g of a substance by 1°C; for water/dilute solutions c ≈ 4.18 J g⁻¹ °C⁻¹.
ΔT (temperature change)
T(final) − T(initial); positive if temperature rises (exothermic reaction); negative if temperature falls (endothermic).
Systematic error in combustion calorimetry
Heat lost to the surroundings, incomplete combustion, and evaporation of fuel from the wick between weighings, which usually reduce the measured |ΔHc|.
Cross-lesson links: Combustion calorimetry (this lesson) is one of three measurement methods in this module. L03 applies the same q = mcΔT logic to neutralisation reactions; L04 applies it to dissolution of ionic solids. L06 and L07 show you how to calculate ΔHc without running an experiment, using bond energies and standard enthalpies of formation respectively. L10 brings all three methods together and asks you to compare fuels using each.
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The Spirit Burner Calorimeter Setup
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The Spirit Burner Calorimeter Setup
Must knowcore concept
The spirit burner calorimeter is a simple, imperfect device, understanding its setup is inseparable from understanding why experimental ΔHc values are usually lower than the accepted value.
Spirit burner calorimeter. The copper cup holds a known mass of water. The fuel burns below, transferring heat upward. The draught shield reduces but does not eliminate heat loss to the surroundings.
Components: spirit burner containing a school-approved fuel (ethanol is the usual choice), a copper calorimeter holding a known mass of water, a thermometer with its bulb fully submerged, and a draught shield around the apparatus.
Safety, supervised school practical only. This is a laboratory activity run under teacher supervision against the school's approved risk assessment, not a home experiment. Wear safety glasses and tie back long hair; keep flammable fuels stoppered and well away from the flame; burn only small quantities with good ventilation; and keep a fire blanket or sand bucket nearby. Use ethanol where possible, methanol is toxic (harmful if inhaled, swallowed or absorbed through skin) as well as highly flammable, so it should only be used if the school risk assessment specifically allows it.
Procedure (in order):
Record initial mass of burner + fuel (balance reading before)
Record initial water temperature
Light the burner; burn fuel until a target temperature rise is reached (e.g. ΔT ≈ 10°C)
Extinguish the flame; record final water temperature
Record final mass of burner + fuel (balance reading after)
Calculate: mass of fuel burned = initial mass − final mass
Copper is chosen for the calorimeter because it has high thermal conductivity heat transfers quickly from the flame to the water, minimising lag between the flame and the thermometer reading.
Weigh the burner before AND after burning. Never estimate or assume the mass of fuel consumed. Evaporation from the wick between weighings adds to the apparent mass lost, always cap the burner immediately after extinguishing the flame to minimise this error.
Common error, not capping the burner: If the burner is left uncapped after the experiment, fuel evaporates from the wick. This makes the recorded mass of fuel burned appear larger than it actually was, giving a falsely small value of n, and therefore a falsely low (less negative) ΔHc.
Going deeper: how the accepted value is measured (bomb calorimeter)
Bomb calorimeter vs spirit burner: Accepted ΔHc values come from a sealed bomb calorimeter, a far more controlled system. Because a bomb operates at constant volume, it directly measures the heat at constant volume (related to the internal-energy change ΔU); a small thermodynamic correction converts this to the constant-pressure enthalpy of combustion ΔHc. Open-air spirit burner experiments lose significant heat, so they usually give values below the accepted figure.
Trace the signs and the split: the reaction system loses heat, the water gains heat, and a real apparatus does not transfer all released energy into the measured water.
CALORIMETRY CALCULATOR, INTERACTIVEInteractive
Adjust mass, specific heat capacity, and ΔT sliders, see q and ΔH update in real time.
In combustion calorimetry, q = mcΔT (m = mass of water, c = 4.18 J g⁻¹ K⁻¹); then ΔHc = −q/n (n = moles of fuel). This simple energy balance assumes all the heat reaches the water and neglects the calorimeter's heat capacity, heat loss and incomplete combustion, so it is an experimental approximation: the dominant systematic errors usually make the measured ΔHc less negative than the accepted value.
Pause, copy the highlighted definition into your book before moving on.
Mini-task: A student sets up a spirit burner experiment but forgets to cap the burner after extinguishing the flame. Will this make their calculated ΔHc too high, too low, or unchanged? Explain the chain of reasoning in two sentences.
Fuel evaporates from the uncapped wick, increasing the apparent mass of fuel burned. This gives a larger-than-true n, so ΔHc = −q/n will be less negative (too low in magnitude) than the true value.
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Sources of Error in Combustion Calorimetry
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Sources of Error in Combustion Calorimetry
Should knowcore concept
We just saw the spirit burner setup and how ΔHc = −q/n is calculated. That raises a question: why does the experimental ΔHc usually come out less negative than the textbook value? This card answers it → by identifying the specific systematic errors that reduce the measured heat.
In a school spirit burner the measured ΔHc is usually less negative than the true value; the key skill is identifying exactly why, and stating the direction of each error's effect.
The experimental ΔHc is usually less negative (lower in magnitude) than the accepted value because the dominant systematic errors all reduce the heat reaching the water; random measurement errors, by contrast, can shift a result in either direction. In HSC questions, you must name the specific physical source of error and state its directional effect on the result.
Effect on ΔHc
q too small → |ΔHc| too low
Less energy released → |ΔHc| too low
m(fuel) overestimated → n too large → |ΔHc| too low
q underestimated (not all heat goes to water) → |ΔHc| too low
ΔT underestimated → q too small → |ΔHc| too low
How to Minimise
Draught shield; insulation around calorimeter
Adequate air supply; clean wick
Cap burner immediately after extinguishing
Use thin-walled calorimeter; correct for calorimeter heat capacity
Ensure thermometer bulb is fully submerged in water
Going deeper: real-world fuel energy density (sets up Short Answer Q3)
Real-World Anchor, Fuel Energy Density: Aviation kerosene (jet fuel) is chosen for aircraft for several reasons, including its freezing point, volatility, flash point and the needs of jet engines, not because it stores much more energy per kilogram than petrol (their gravimetric energy densities are broadly similar). Calorimetry at industrial scale is how fuel engineers compare and certify fuel grades before use. The spirit burner experiment you're running in class is conceptually identical, just less precise. You'll compare fuel energy densities in Short Answer Q3.
Never write "human error" or "inaccurate measurements" in an HSC answer. On their own, these vague phrases usually earn no marks. Always name the specific physical process causing the error (e.g. "heat loss to the surrounding air through the walls of the apparatus") and state its directional effect on ΔHc.
These systematic errors usually make a spirit burner ΔHc less negative than the accepted value. Five named sources: (1) heat loss to surroundings, (2) incomplete combustion, (3) fuel evaporation (uncapped burner), (4) heat absorbed by copper walls, (5) thermometer not submerged. Each reduces |ΔHc|; remember that random measurement errors can scatter either way.
Add the highlighted point to your notes before the check below.
Explain it: In your own words, explain why a school spirit burner usually gives a ΔHc that is less negative than the accepted value. Use the phrase "systematic error" and name at least two specific physical sources.
Model answer: In a spirit burner experiment, ΔHc is usually less negative than the accepted value because the dominant systematic errors reduce the heat measured. The two main sources are: (1) heat loss to the surrounding air, combustion energy heats the air rather than the water, making q smaller; (2) incomplete combustion, some fuel produces CO instead of CO₂, releasing less energy per mole. Both errors make |ΔHc| smaller than the true value.
Same science, choose your support
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Same science, choose your support
pick one
All three routes practise the same Must-Know skill: turn calorimetry data into a heat value with q = mcΔT. Do one; try another if you want more practice.
Supported word bank + sentence frame start here if unsure
Word bank: mass of water · 4.18 · temperature rise · positive
A spirit burner heats 150.0 g of water, and the temperature rises by 9.0 K. Complete the frame:
"q = m × c × ΔT = 150.0 g × ______ J g⁻¹ K⁻¹ × ______ K = ______ J. Because the water gains heat, q is ______."
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Show a hint
Use the mass of water (not fuel) as m, take c = 4.18, and multiply by the 9.0 K rise. The answer is 150.0 × 4.18 × 9.0 = 5643 J, and q is positive because the water absorbs heat.
Core independent HSC-standard the exam standard
A student burns 0.68 g of ethanol (C₂H₅OH, M = 46.07 g mol⁻¹), and 200.0 g of water rises from 20.4 °C to 32.1 °C. Calculate q in kJ, then ΔHc in kJ mol⁻¹. State the sign and what it tells you about the reaction.
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Stretch reasoning in a new context push yourself
Two students burn the same fuel and both calculate q correctly. One caps the burner immediately; the other leaves it uncapped for a minute before the final weighing. Predict whose ΔHc is less negative, explain the chain of reasoning using n and ΔHc = −q/n, and suggest one change to the apparatus that would bring both closer to the accepted value.
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Worked examples
Worked examples · reveal as you go
Worked example 1+5 XP on full reveal
Calculating q and ΔHc. A student burns ethanol (C₂H₅OH, M = 46.07 g mol⁻¹) in a spirit burner. The mass of the burner decreases from 152.34 g to 151.62 g. The calorimeter contains 200.0 g of water. The water temperature rises from 19.5°C to 31.2°C. Calculate the molar enthalpy of combustion of ethanol.
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Mass of fuel burned m(fuel) = 152.34 − 151.62 = 0.72 g
Subtract final burner mass from initial. This is the mass of ethanol consumed.
The accepted value is −1367 kJ mol⁻¹. The experimental value is less than half, showing very significant heat loss, typical of spirit burner experiments.
ΔHc = −q/n. q must be in kJ. The negative sign reflects that combustion is exothermic.
Worked example 2+5 XP on full reveal
Percentage error & sources of discrepancy. The accepted molar enthalpy of combustion of methanol (CH₃OH) is −726 kJ mol⁻¹. A student's experiment gave −412 kJ mol⁻¹. (a) Calculate the percentage error. (b) Suggest two specific sources of error that would explain why the experimental value is less negative.
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Set up percentage error formula $\% \text{ error} = \dfrac{|\text{experimental} - \text{accepted}|}{|\text{accepted}|} \times 100$
Use absolute values, the sign of ΔH does not affect percentage difference.
A 43% error is large but typical for an open spirit burner experiment, most of the combustion energy escapes to the surroundings before heating the water.
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Two specific sources of error (1) Heat loss to surroundings: A large proportion of combustion energy heats the surrounding air and copper walls rather than the water, making q smaller than the true heat released, giving a ΔHc less negative than the accepted value.
(2) Incomplete combustion: Insufficient oxygen at the flame produces CO instead of CO₂ for some of the combustion. CO releases less energy per mole than complete combustion to CO₂, reducing the total heat released and making ΔHc less negative.
Each source must name the physical process and state its directional effect. "Human error" on its own usually earns no marks.
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Final Answer (a) 43.3% error. (b) Heat loss to surroundings; incomplete combustion. Both make ΔHc less negative than the accepted value.
Full marks in HSC require both a named physical source and its directional effect on ΔHc for each error identified.
Key idea
Sort the steps+7 XP
Put the calorimetry combustion procedure in the correct order. Click two steps to swap them, then click Check order.
Calculate q = mcΔT for the water in the calorimeter.
Measure the initial mass of the fuel + spirit burner.
Divide q by the moles of fuel burned to get ΔHcomb.
Record the initial and final temperature of the water.
Find the mass of fuel burned (initial mass − final mass).
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What you'll master, and the words for it
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Formula reference · this lesson
core formula
📐
Formula Reference, This Lesson
$q = mc\Delta T$
q = heat energy absorbed by water (J)m = mass of water in calorimeter (g)c = specific heat capacity of water = 4.18 J g⁻¹ K⁻¹ΔT = temperature change (°C or K, same magnitude)
$\Delta H_c = \dfrac{-q}{n}$
ΔHc = molar enthalpy of combustion (kJ mol⁻¹)q = heat energy in kJ (divide J answer by 1000)n = moles of fuel burnedNegative sign: combustion is exothermic, q is positive but ΔHc is negative
$n = \dfrac{m}{M}$
n = moles of fuel burned (mol)m = mass of fuel burned (g) = initial burner mass − final burner massM = molar mass of fuel (g mol⁻¹)
Full calculation chain:
mass of fuel burned → n (using n = m/M) → q (using q = mcΔT, m = water mass) → ΔHc (using ΔHc = −q/n, with q in kJ)
Warm up and recall
Printable worksheet
Download this lesson's worksheet
Use the PDF for classwork, homework or revision. It includes key ideas, activities, questions, an extend task and success-criteria proof.
Students substitute the mass of fuel burned into q = mcΔT instead of the mass of water in the calorimeter.
Fix: The 'm' in q = mcΔT is always the mass of the substance being heated, in a combustion calorimeter, that is the water. The fuel is the energy source, not the substance whose temperature you're measuring. q tells you how much energy the water absorbed, and c is the specific heat capacity of water.
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"The negative sign in ΔHc = −q/n is a mistake"
Students omit the negative sign or write ΔHc = +q/n, giving a positive value for combustion.
Fix: Combustion is exothermic, ΔHc must be negative. The formula ΔHc = −q/n converts the positive q (heat absorbed by water) into the correct negative enthalpy change for the reaction. If you get a positive ΔHc for combustion, you've made a sign error.
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"Human error explains the discrepancy"
Students write "human error" or "inaccurate measurements" as sources of error in an HSC answer.
Fix: On their own, these vague phrases usually earn no marks. Always name the specific physical process (e.g. "heat loss to the surrounding air through the open sides of the apparatus") and state its directional effect on ΔHc (e.g. "this makes q smaller than the true heat released, so ΔHc is less negative than the accepted value").
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Drill, then revisit
Quick-fire practice · 5 reps +2 XP per reveal
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Calorimetry Data Processing, Propan-1-ol: A student burns propan-1-ol (C₃H₇OH, M = 60.10 g mol⁻¹). Burner mass: before 188.45 g, after 187.81 g. Water mass: 250.0 g. Temperature: initial 18.2°C, final 32.6°C.
(a) Calculate mass and moles of propan-1-ol burned.
(b) Calculate q in kJ.
(c) Calculate ΔHc and the percentage error (accepted value = −2021 kJ mol⁻¹).
Error analysis, butan-1-ol: A student burns 0.80 g of butan-1-ol (C₄H₉OH, M = 74.12 g mol⁻¹) and heats 200.0 g of water from 18.0°C to 33.0°C (ΔT = 15.0°C). Scenario 1, wrong m: The student uses m = 0.80 g (mass of fuel) instead of 200.0 g in q = mcΔT. Calculate the incorrect q and ΔHc. Scenario 2, evaporation: The burner was not capped; recorded mass loss is 0.90 g instead of 0.80 g. Calculate the resulting ΔHc.
Scenario 2 (evaporation): n(wrong) = 0.90/74.12 = 0.01214 mol; q unchanged = 12.54 kJ; ΔHc(wrong) = −12.54/0.01214 = −1033 kJ mol⁻¹. Less negative than correct −1162, overestimated n means smaller |ΔHc|.
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Short calculation: A student burns 0.500 g of an unknown fuel (M = 100 g mol⁻¹) and records a temperature rise of 8.4°C in 150.0 g of water. Calculate q in kJ and then ΔHc in kJ mol⁻¹.
Sign and direction reasoning: Water absorbs heat during combustion calorimetry, so q (for the water) is positive. The combustion reaction releases heat, so ΔH (for the reaction) is negative. A student writes ΔHc = +q/n. Identify the error and explain the correct sign convention using the formula ΔHc = −q/n.
The error: using +q/n gives a positive ΔHc, which would incorrectly indicate an endothermic reaction. Water absorbs heat (q > 0, positive for the water), but the reaction releases that heat, so ΔH for the reaction must be negative. The formula ΔHc = −q/n converts the positive q into the correct negative enthalpy change. Combustion always gives ΔHc < 0.
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Extension, systematic error direction: A student's spirit burner experiment consistently gives ΔHc values about 50% below the accepted NESA data sheet values. A classmate suggests the balance is imprecise and introduces 0.01 g random error in the fuel mass reading. Could balance imprecision explain the 50% shortfall? Explain your reasoning.
No, balance imprecision would produce random scatter (some results too high, some too low), not a consistent 50% shortfall always in the same direction. The consistent direction of error, always less negative, points to a systematic source. The dominant systematic error in spirit burner experiments is heat loss to the surrounding air, which consistently reduces the measured q and therefore |ΔHc|. A 50% shortfall is entirely consistent with heat loss, not balance precision.
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Revisit your thinking
Go back to your Think First response. Now that you've studied combustion calorimetry, return to Lavoisier and Laplace's 1780 ice calorimeter at the French Academy of Sciences:
Their ice calorimeter measured mass of meltwater rather than ΔT. How does this differ from the spirit burner method? Write the formula they would have used, relating heat to mass of ice melted and the latent heat of fusion (334 J g⁻¹).
Lavoisier and Laplace got ~34 kJ g⁻¹ for charcoal. If you burned 1.50 g of charcoal in a spirit burner and heated 200 g of water, what ΔT would you expect if no heat was lost? Show your working.
Their result was a lower bound on the true value, just like your spirit burner result. What was their main source of heat loss, and what was yours?
Use the Calorimetry Calculator. Heating 200 g of water by 5.0°C (c = 4.18 J/g·°C) requires how much heat?
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Multiple choice
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Multiple choice
+5 XP per correct · +25 XP all-correct
Pick your answer, then rate your confidence. That tells the system what to drill next.
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Misconceptions to fix before short answer
Wrong: Incomplete combustion is safer than complete combustion because it produces less CO₂.
Right: Incomplete combustion produces toxic carbon monoxide (CO) and particulate carbon (soot), which are immediately dangerous to health. Complete combustion produces CO₂ and H₂O, which are far less acutely toxic, though CO₂ is an asphyxiant at high concentration and a greenhouse gas. We still aim for efficient, complete combustion for safety and efficiency.
Wrong: "m in q = mcΔT is the mass of fuel burned."
Right: m is the mass of water in the calorimeter, the substance being heated. The specific heat capacity c = 4.18 J g⁻¹ °C⁻¹ is the value for water, not the fuel.
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Short answer
UnderstandBand 3
Q1. Explain why, in combustion calorimetry using a spirit burner, the value of 'm' in the formula q = mcΔT must be the mass of water and not the mass of fuel. In your answer, state what q represents and why the specific heat capacity of water is used. 3 MARKS
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ApplyBand 4
Q2. A student burns 0.85 g of methanol (CH₃OH, M = 32.04 g mol⁻¹) and heats 180.0 g of water from 21.0°C to 34.5°C. (a) Calculate q in kJ. (b) Calculate ΔHc. (c) The accepted value is −726 kJ mol⁻¹. Calculate the percentage error and identify one specific source of error that explains why the experimental value is less negative. 5 MARKS
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EvaluateBand 5
Q3.Real-World Application: Fuel engineers compare fuels by their molar enthalpy of combustion (ΔHc) per gram (energy density). The table below shows data for three fuels.
Fuel
Formula
M (g mol⁻¹)
ΔHc (kJ mol⁻¹)
Energy per gram (kJ g⁻¹)
Methanol
CH₃OH
32.04
−726
Calculate
Ethanol
C₂H₅OH
46.07
−1367
Calculate
Octane (petrol proxy)
C₈H₁₈
114.23
−5471
Calculate
(a) Complete the energy per gram column (kJ g⁻¹ = |ΔHc| ÷ M). (3 marks)
(b) Using your calculated values, explain which fuel is most efficient per gram and suggest why octane is preferred in internal combustion engines over methanol. (2 marks)5 MARKS
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Comprehensive Answers
Show comprehensive answers ▼
Short Answer Model Answers
Q1 (3 marks): q represents the heat energy absorbed by the water during combustion [1]. 'm' must be the mass of water because q = mcΔT calculates the thermal energy transferred to the water, not to the fuel or the calorimeter [1]. The specific heat capacity of water (c = 4.18 J g⁻¹ K⁻¹) is used because water is the substance being heated, and c tells us how many joules are needed to raise 1 g by 1 K [1].
Q2 (5 marks): (a) ΔT = 34.5 − 21.0 = 13.5°C; q = 180.0 × 4.18 × 13.5 = 10,157 J = 10.157 kJ [1]. (b) n = 0.85/32.04 = 0.02653 mol; ΔHc = −10.157/0.02653 = −383 kJ mol⁻¹ [2-1 for n, 1 for ΔHc]. (c) % error = |−383 − (−726)| / 726 × 100 = 343/726 × 100 = 47.2% [1]. Source: heat loss to surroundings, a large proportion of the combustion energy heats the surrounding air rather than the water, making q smaller than the true heat released and ΔHc less negative [1].
Q3 (5 marks): (a) Methanol: 726/32.04 = 22.7 kJ g⁻¹ [1]; Ethanol: 1367/46.07 = 29.7 kJ g⁻¹ [1]; Octane: 5471/114.23 = 47.9 kJ g⁻¹ [1]. (b) Octane is most efficient per gram at 47.9 kJ g⁻¹, more than twice methanol's 22.7 kJ g⁻¹ [1]. Octane (petrol) is preferred in internal combustion engines because it releases more energy per gram of fuel, allowing vehicles to travel further on the same tank mass, this is the same reason jet fuel (kerosene) is denser in energy than wood or alcohol [1].
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Check what actually stuck
Check what actually stuck
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Boss battle
earn bronze · silver · gold
Five timed questions on Calorimetry, Combustion. Beat the boss to bank a tier, gold (perfect + fast), silver (80%+), or bronze (cleared).
