Lesson 3-1c · Calorimetry & Energy Systems

You cannot create energy. But you can trap it.

How do scientists measure the Calories in an almond? You cannot stick an energy meter into a molecule. Instead, you build a thermal trap called a calorimeter, an insulated water bath that absorbs the energy the sample gives up. The water’s temperature change then tells you how much energy moved.

Alignment
HS-PS3-1 Create a computational model to calculate the change in the energy of one component in a system when the change in energy of the other component(s) and energy flows in and out of the system are known.
Objective
You can define the system and the surroundings and show that qsystem + qsurroundings = 0. You can draw an LOL energy bar diagram. You can calculate an unknown specific heat or a final temperature.
Scope
System and surroundings, coffee-cup calorimetry, qlost = −qgained, and LOL energy accounting.

Core Claims

  • Energy is conserved. In an isolated container the energy changes add to zero: ΔEsystem + ΔEsurroundings = 0.
  • System and surroundings. The system is the sample or reaction you study. The surroundings are the water bath, the cup, and the thermometer that exchange energy with it.
  • The conservation equation. Heat lost by the hotter part equals heat gained by the cooler part: qlost = −qgained, or qmetal = −qwater.
  • Signs. Cooling gives a negative ΔT and a negative q. Warming gives a positive ΔT and a positive q.
  • LOL energy bar diagrams. An LOL diagram shows the initial energy (L), the transfers across the boundary (O), and the final energy (L). The total stays the same.

Calorimeter boundary

Surroundings (water) System hot metal q q
heat q leaves the system and enters the surroundings

Retrieval Checklist

  • Name the system and the surroundings in a calorimeter.
  • State that energy is conserved and write the conservation equation.
  • Explain why qsystem = −qsurroundings needs a negative sign.
  • Draw an LOL diagram that tracks thermal energy (Eth).

Where you draw the boundary changes everything.

To track energy, you must draw a clear border. Thermodynamics splits the universe into two parts.

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The system

The reaction, the dissolving salt, or the heated metal that you chose to study. In coffee-cup calorimetry, the system is often a hot metal cylinder.

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The surroundings

Everything else that can exchange energy with the system. In the lab, that means the water bath, the foam cups, the thermometer, and the stirring rod.

The foundation of thermochemistry is the First Law of Thermodynamics: energy cannot be created or destroyed. It only moves between components or changes form.

qsystem + qsurroundings = 0
qsystem = −qsurroundings

Heat released by the system (−q) is absorbed by the surroundings (+q). The total thermal energy in an isolated calorimeter never changes.

Common mistake

Expecting equal masses to give equal temperature changes. 50 g of metal and 50 g of water swap equal amounts of energy, but not equal numbers of degrees. Water has a high specific heat, so its temperature changes far less.

The coffee-cup calorimeter: cheap, simple, and accurate.

Why do chemistry classes use foam coffee cups to measure energy? Expanded polystyrene is over 95% trapped, motionless air. That air insulates well, so little heat leaks to the room by conduction or convection.

Nested cup Insulating air gap Water bath (surroundings) Metal Foam lid stops evaporation Thermometer Glass stirrer The governing balance q_metal = −q_water Measure the water: q_w = m_w × c_w × ΔT_w Then find the metal: c_m = −q_w / (m_m × ΔT_m)
Figure 3.4. two nested foam cups and a tight lid trap air, so heat moves only between the metal (system) and the water (surroundings)

A coffee-cup experiment needs four measurements and one calculation.

  1. Metal mass (mm) and starting temperature (Tm,i). Heat the metal in boiling water to a known temperature, usually 100.0 °C.
  2. Water mass (mw) and starting temperature (Tw,i). Weigh cool water into the calorimeter.
  3. Final temperature (Tf). Move the hot metal into the calorimeter fast and close the lid. Stir gently until the thermometer peaks.
  4. Calculation. The metal and the water end at the same Tf. Solve for the unknown specific heat:
    cmetal = −[mw × cw × (Tf − Tw,i)] / [mm × (Tf − Tm,i)]
Common mistake

Using two different final temperatures. The metal and the water end at the same Tf. Only their starting temperatures differ.

LOL diagrams: visual accounting for energy storage.

The equation qlost = −qgained shows the balance in numbers. It does not show where the energy lives. Teachers in the Modeling Instruction curriculum developed the LOL energy bar diagram to track energy accounts by sight.

Initial energy (L) 1 2 3 4 4 blocks Metal 1 block Water Total initial = 5 System (O) Isolated cup q = 2 blocks Metal −2, water +2 ΔE_sys = 0 Final energy (L) 1 2 3 4 2 blocks Metal 3 blocks Water Total final = 5
Figure 3.5. the metal loses 2 blocks and the water gains 2 blocks · the total stays at 5 because nothing crosses the cup wall

An LOL diagram follows three rules.

  • Left “L” (initial energy). Bars show the energy stored in each component before they interact. The hot metal starts with tall thermal energy bars (Eth). The cold water starts with short bars.
  • Center “O” (energy flow circle). The circle is the system boundary. In an insulated calorimeter no heat crosses into or out of the room (Ein = 0, Eout = 0). Energy only moves between the parts inside.
  • Right “L” (final energy). Bars show the energy at thermal equilibrium. The metal lost bars. The water gained the same number. Total initial blocks + Ein − Eout = total final blocks.
Common mistake

Counting energy inside the center circle. The circle is a boundary, not a storage place. The energy sits in the bars on the left and the right.

Coffee-cup calorimeter and LOL workbench.

Pick a metal and set the masses and starting temperatures. Press Drop metal and stir to watch heat move. The LOL bars show the energy blocks balancing.

Simulation

Why does the metal lose more degrees than the water gains?

Water now20.0 °C
Final temperature–
Metal change–
Water change–
Heat exchanged–
Predict to start

The explanation appears after you predict.

  • The metal sample (the system)
  • The water (the surroundings)
  • LOL block: one unit of thermal energy, not to scale
  1. Load equal masses, drop the metal, then press Record. Look for: the metal cools about 73 °C while the water warms about 7 °C.
  2. Load lead, drop it, then press Record. Look for: the final temperature barely moves above 20 °C.
  3. Load aluminum with the same masses, drop it, then press Record. Look for: a much higher final temperature than lead gave.
  4. Load a large iron sample, drop it, then press Record. Look for: a large mass of iron lifts the water to about 44 °C.

Use your evidence

From coffee cups to bomb calorimeters: measuring food Calories.

A coffee-cup calorimeter works well for reactions in water and for metals. It cannot handle combustion. Burn a walnut in a paper cup and the water boils away while smoke escapes.

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The constant-volume bomb calorimeter

To measure the energy in a fuel or a food, scientists use a heavy, sealed steel container called a bomb calorimeter. The sample sits in a crucible with pure oxygen at about 30 atm. Electric wires ignite it, and it burns completely in a fraction of a second. The heat flows into a surrounding water jacket.
qcombustion = −Ccalorimeter × ΔT

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What is a dietary “Calorie”?

Nutrition labels list “Calories” with a capital C. In chemistry, 1 calorie (cal) is the energy that warms 1 g of water by 1 °C (4.184 J). A dietary Calorie (Cal) is a kilocalorie: 1,000 cal or 4,184 J. A 250-Calorie candy bar gives your body over 1,000,000 J (1 megajoule) of chemical energy.

Common mistake

Mixing up calorie and Calorie. A Calorie (capital C) is 1,000 calories. A 250-Calorie bar holds 250,000 calories, not 250.

Macronutrient energy densities

Energy released per gram by each macronutrient class, measured by calorimetry
Macronutrient class Dietary energy (Cal/g) Energy (kJ/g) Why
Carbohydrates (sugars, starches) 4.0 17 Partly oxidized, with many C–O and O–H bonds already present.
Proteins (polypeptides) 4.0 17 Similar to carbohydrates. Excreting nitrogen as urea uses some energy.
Fats (triglycerides) 9.0 38 Highly reduced hydrocarbon tails hold over twice the energy per gram.

Fill the blanks from memory.

Type each blank from memory, then press Check. Press Show answers after your first check if you are stuck.

1. In an isolated system, the total energy is .
2. The calorimetry equation is qsystem = .
3. In a coffee-cup calorimeter, the water is the and the metal or reacting chemicals are the system.
4. In an LOL diagram, the center letter “O” stands for the , where energy flows.
5. One dietary Calorie (capital C) equals calories of heat.

A student drops a 100.0 g block of aluminum at 90.0 °C into 100.0 g of water at 20.0 °C in a coffee-cup calorimeter.

The metal cools by about 58 °C. The water warms by only about 12 °C. The student asks: “If energy is conserved, why did the metal lose 58 degrees while the water gained only 12? Did energy disappear?”

Write 3 to 4 sentences that settle the confusion. Use specific heat and energy conservation.

Separate the temperature change (ΔT) from the thermal energy transferred (q). Write your answer first, then compare it with the model answer.

One way to say it

The student is mixing up the temperature change (ΔT) with the energy transferred (q). Conservation says the metal loses exactly the energy the water gains: −qmetal = qwater. It does not say the temperature changes are equal.

Because q = mcΔT, the temperature change is ΔT = q / mc. Water has a specific heat of 4.184 J/(g·°C), more than four times that of aluminum (0.897 J/(g·°C)). The same energy changes the water’s temperature by far fewer degrees than the metal’s. No energy was created or destroyed.

Write your answer first. Then grade yourself.

Give yourself 1 mark for each idea you wrote down.

Gen Chem · HS-PS3-1 · constructed response [4 marks]

A chemist runs a coffee-cup calorimetry experiment to identify an unknown metal cylinder.

Data:
• Mass of metal: 65.00 g
• Starting temperature of metal: 99.5 °C
• Mass of water: 120.00 g
• Starting temperature of water: 21.2 °C
• Final temperature of the mixture: 24.8 °C
• Specific heat of water: 4.184 J/(g·°C)

(a) Calculate the heat, in joules, absorbed by the water. Show your work with units. [1 mark]

(b) Assume an ideal isolated calorimeter. Calculate the specific heat of the metal in J/(g·°C). [2 marks]

(c) The experiment is repeated with no lid on the cup. Say whether your calculated specific heat is higher, lower, or the same as the true value. Justify your answer. [1 mark]

Scoring guide: 4 marks
  • Water heat (1 mark): You find qwater = 1,807.5 J (accept 1,810 J).
  • Conservation (1 mark): You set qmetal = −qwater and find ΔTmetal = −74.7 °C.
  • Specific heat (1 mark): You calculate c = 0.372 J/(g·°C) (accept 0.37 to 0.38).
  • Error direction (1 mark): You say c comes out lower, because heat escapes and Tf is too low.
Why this matters

Electric vehicle batteries make a lot of waste heat during fast acceleration and fast charging. Engineers measure the heat a battery cell gives off (−qbattery). Then they design a liquid coolant loop that absorbs it (+qcoolant). The loop keeps the battery pack from overheating.

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Adaptive Competency Practice: HS-PS3-1

Test your understanding of calorimetry and energy conservation balances. The adaptive engine models coffee-cup calorimetry (q_sys + q_surr = 0), solution molar enthalpy calculations, LOL bar charts, and experimental error propagation.

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