Thermochemistry

CHEM 50 — General Chemistry I · Dr. Karmach

build Jul 13 · 16:18 · CC BY-NC-SA 4.0 · OpenStax Chemistry 2e where noted
Dr. Karmach

By the end of this unit, you can…

  • Distinguish heat from work and convert among joules, calories, and Calories
  • Identify exothermic and endothermic processes from equations, energy diagrams, and the sign of ΔH
  • Apply q = m·c·ΔT to find heat, mass, specific heat, or temperature change
  • Compare how specific heats set different substances' temperature response to the same heat
  • Use calorimetry data to determine the heat of a process or reaction
  • Reverse and scale thermochemical equations and use ΔH as a conversion factor
Dr. Karmach

Today's route 🗺️

  1. Energy & Its Units
  2. Exothermic & Endothermic
  3. Specific Heat & q = mcΔT
  4. Comparing Specific Heats
  5. Calorimetry
  6. Thermochemical Equations
Dr. Karmach

1 · Energy & Its Units

Tell kinetic from potential energy and heat from temperature, and convert any amount of energy among joules, calories, kilojoules, and food Calories.

Dr. Karmach

Same snack, two labels

The same snack bar sells in two countries. One label lists 230 Calories; the other lists 960 kJ. Both describe the same energy, counted in different units.

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Energy is never created or destroyed

Energy is the capacity to transfer heat or to do work. A process moves energy from one place to another or changes its form. The total amount never changes.

energy leaving one place = energy arriving in another
the total is fixed — so an amount of energy can be counted, like mass
Dr. Karmach

Kinetic and potential energy

Kinetic energy is energy of motion. Potential energy is energy stored by position or arrangement. Fuels hold chemical potential energy — stored in the arrangement of atoms, released when a reaction rearranges them.

Dr. Karmach

Two ways energy transfers: heat and work

Energy transfers between things in exactly two ways. Heat (q) is transfer driven by a temperature difference. Work (w) is transfer by a force moving something.

heat (q): hot pan → cool water
energy flows because the temperatures differ
work (w): expanding gas pushes a piston
energy moves because a force acts through a distance
Dr. Karmach

Temperature is not heat

Temperature measures the average kinetic energy of the particles: an intensity. Heat is an amount of energy in transfer. More sample means more energy at the same temperature.

a cup and a bathtub, both at 40 °C
same temperature — the tub transfers far more heat as it cools
Dr. Karmach

The units of energy

The SI unit is the joule (J). 1 cal = 4.184 J, exactly. The food Calorie has a capital C: 1 Cal = 1 kcal = 1000 cal. Each equality is an ordinary conversion factor.

Dr. Karmach

The method

  1. Write the given: number and unit.
  2. Plan the route: given unit → wanted unit. Read a capital C as 1000 cal.
  3. Chain the factors so each unit cancels.
  4. Sense-check the size and the surviving unit.
Dr. Karmach

Worked example 1 — calories to joules

Step 1 · Write the given

1 cal = 4.184 J
given: 175 cal · wanted: J

A single-use hand warmer releases 175 cal of heat as the iron inside it oxidizes. Express the energy in joules.

Dr. Karmach

Worked example 1 — solution

1 cal = 4.184 J
given: 175 cal · wanted: J

Step 2 · Plan the route

One arrow links the units: cal → J. One conversion factor is needed.

Dr. Karmach

Worked example 1 — solution

1 cal = 4.184 J
given: 175 cal · wanted: J
Step 2 · Plan the route Step 3 · Chain the factors so each unit cancels

The factor takes cal in the denominator, so cal cancels and J survives:

175 cal × 4.184 J1 cal = 732 J
Dr. Karmach

Worked example 1 — solution

1 cal = 4.184 J
given: 175 cal · wanted: J
Step 2 · Plan the route Step 3 · Chain the factors so each unit cancels
175 cal × 4.184 J1 cal = 732 J
Step 4 · Sense-check
A joule is smaller than a calorie, so the count in joules must be larger: 175 → 732. The heat itself is unchanged. ✓
Dr. Karmach

Worked example 2 — kilojoules to calories

Step 1 · Write the given

1 kJ = 1000 J · 1 cal = 4.184 J
given: 2.50 kJ · wanted: cal

An instant cold pack absorbs 2.50 kJ of heat from the skin it touches. Express the energy in calories.

Dr. Karmach

Worked example 2 — solution

1 kJ = 1000 J · 1 cal = 4.184 J
given: 2.50 kJ · wanted: cal · plan: kJ → J → cal

Step 2 · Plan the route

No single equality links kJ to cal. The route runs through the joule: kJ → J → cal. Two conversion factors are needed.

Dr. Karmach

Worked example 2 — solution

1 kJ = 1000 J · 1 cal = 4.184 J
given: 2.50 kJ · wanted: cal · plan: kJ → J → cal
Step 2 · Plan the route Step 3 · Chain the factors so each unit cancels

The prefix factor cancels kJ; the calorie factor takes J in the denominator:

2.50 kJ × 1000 J1 kJ × 1 cal4.184 J = 598 cal
Dr. Karmach

Worked example 2 — solution

1 kJ = 1000 J · 1 cal = 4.184 J
given: 2.50 kJ · wanted: cal · plan: kJ → J → cal
Step 2 · Plan the route Step 3 · Chain the factors so each unit cancels
2.50 kJ × 1000 J1 kJ × 1 cal4.184 J = 598 cal
Step 4 · Sense-check
One calorie holds 4.184 J, so 2500 J make fewer calories than joules: 598. The pack absorbs the same heat under either name. ✓
Dr. Karmach

Your turn — calories to kilojoules

1 cal = 4.184 J · 1 kJ = 1000 J
given: 7.10 × 10³ cal · wanted: kJ · plan: cal → J → kJ

Burning one gram of ethanol releases 7.10 × 10³ cal.

7.10 × 10³ cal × 4.184 J1 cal × kJ J = kJ

Fill the second factor from 1 kJ = 1000 J, then compute.

Dr. Karmach

Your turn — calories to kilojoules

1 cal = 4.184 J · 1 kJ = 1000 J
given: 7.10 × 10³ cal · wanted: kJ · plan: cal → J → kJ

Burning one gram of ethanol releases 7.10 × 10³ cal.

7.10 × 10³ cal × 4.184 J1 cal × kJ J = kJ

Fill the second factor from 1 kJ = 1000 J, then compute.

7.10 × 10³ cal × 4.184 J1 cal × 1 kJ1000 J = 29.7 kJ
Dr. Karmach

Where this goes wrong

Writing the 4.184 factor upside down. 175 cal × (1 cal / 4.184 J) = 41.8 cal²/J. No unit cancels, and the answer is not in joules. The factor that cancels cal gives 732 J.
Stopping at joules. The plan cal → J → kJ has two arrows. Stopping after one leaves 2.97 × 10⁴ J — joules, not the wanted kilojoules. The chain ends at 29.7 kJ.
Reading temperature as an amount of energy. A cup of tea and a bathtub of water can both read 40 °C. The temperatures match; the tub holds far more energy. Temperature is an intensity; heat is an amount.
Dr. Karmach

Practice 1

1 kJ = 1000 J · 1 cal = 4.184 J
given: 6.20 kJ · wanted: cal

Dissolving calcium chloride in a beaker of water releases 6.20 kJ of heat. How many calories is that?

  1. 2.59 × 10⁴ cal
  2. 1.48 × 10³ cal
  3. 6.20 × 10³ cal
  4. 1.48 cal
Dr. Karmach

Practice 1 — answer: B

1 kJ = 1000 J · 1 cal = 4.184 J
given: 6.20 kJ · plan: kJ → J → cal
6.20 kJ × 1000 J1 kJ × 1 cal4.184 J = 1.48 × 10³ cal — answer B

A flipped the 4.184 factor: 6200 × 4.184 = 2.59 × 10⁴, and no unit cancels. C stopped after the prefix factor: 6.20 × 1000 = 6200, a count of joules, not calories. D treated kilojoules as joules: 6.20 / 4.184 = 1.48, a thousand times too small.

Each calorie holds 4.184 J, so 6200 J make fewer calories than joules: 1.48 × 10³. ✓
Dr. Karmach

Worked example 3 — the food Calorie

Step 1 · Write the given

1 Cal = 1000 cal · 1 cal = 4.184 J · 1 kJ = 1000 J
given: 230 Cal · wanted: kJ

A snack bar's label lists 230 Calories; the same bar abroad is labeled 960 kJ. Express 230 Cal in kilojoules.

A common first attempt treats 230 Calories as 230 calories. Test it.

Dr. Karmach

Worked example 3 — solution

1 Cal = 1000 cal · 1 cal = 4.184 J · 1 kJ = 1000 J
given: 230 Cal · wanted: kJ

A common first attempt

230 cal × 4.184 J1 cal × 1 kJ1000 J = 0.962 kJ ✗
Dr. Karmach

Worked example 3 — solution

1 Cal = 1000 cal · 1 cal = 4.184 J · 1 kJ = 1000 J
given: 230 Cal · wanted: kJ

A common first attempt

230 cal × 4.184 J1 cal × 1 kJ1000 J = 0.962 kJ ✗
The kilojoule label reads 960 — this result is 1000 times too small. The label's unit is Cal, not cal. ✗
Dr. Karmach

Worked example 3 — the correct chain

1 Cal = 1000 cal · 1 cal = 4.184 J · 1 kJ = 1000 J
given: 230 Cal · wanted: kJ

Step 2 · Plan the route

The capital C marks the food Calorie: 1 Cal = 1000 cal. The route: Cal → cal → J → kJ. Three conversion factors are needed.

Dr. Karmach

Worked example 3 — the correct chain

1 Cal = 1000 cal · 1 cal = 4.184 J · 1 kJ = 1000 J
given: 230 Cal · wanted: kJ
Step 2 · Plan the route Step 3 · Chain the factors so each unit cancels
230 Cal × 1000 cal1 Cal × 4.184 J1 cal × 1 kJ1000 J = 962 kJ
Dr. Karmach

Worked example 3 — the correct chain

1 Cal = 1000 cal · 1 cal = 4.184 J · 1 kJ = 1000 J
given: 230 Cal · wanted: kJ
Step 2 · Plan the route Step 3 · Chain the factors so each unit cancels
230 Cal × 1000 cal1 Cal × 4.184 J1 cal × 1 kJ1000 J = 962 kJ
Step 4 · Sense-check
The kilojoule label reads 960: the chain reproduces it, rounded. The same chain shows 1 Cal = 4.184 kJ. ✓
Dr. Karmach

Take-home: the food Calorie is a kilocalorie

Do: read the capital C as 1000 cal.

230 Cal × (1000 cal / 1 Cal) × (4.184 J / 1 cal) × (1 kJ / 1000 J) = 962 kJ
matches the 960 kJ label ✓

Do not: read Cal as cal. The answer lands 1000 times too small.

230 cal × (4.184 J / 1 cal) × (1 kJ / 1000 J) = 0.962 kJ
1000 times smaller than the label ✗
Dr. Karmach

Practice 2

1 Cal = 1000 cal · 1 cal = 4.184 J · 1 kJ = 1000 J
given: 140 Cal · wanted: kJ

A 12-ounce can of regular soda supplies 140 Calories. Express this energy in kilojoules.

  1. 33.5 kJ
  2. 5.86 × 10⁵ kJ
  3. 586 kJ
  4. 0.586 kJ
Dr. Karmach

Practice 2 — answer: C

1 Cal = 1000 cal · 1 cal = 4.184 J · 1 kJ = 1000 J
given: 140 Cal · plan: Cal → cal → J → kJ
140 Cal × 1000 cal1 Cal × 4.184 J1 cal × 1 kJ1000 J = 586 kJ — answer C

A flipped the 4.184 factor: 140 × 1000 / 4.184 / 1000 = 33.5, and the joule never cancels. B stopped at joules: 140 × 1000 × 4.184 = 5.86 × 10⁵, a count of J, not kJ. D read Cal as cal: 140 × 4.184 / 1000 = 0.586, a thousand times too small.

One food Calorie is 4.184 kJ, so 140 Cal sits near 140 × 4 = 560. The chain gives 586. ✓
Dr. Karmach

Check yourself

  1. A label lists 95 Calories. Write the chain to kilojoules: which factor comes first, and what does each unit cancel into?
  2. A cup of water and a pot of water both read 60 °C. Which quantity matches, and which differs: temperature, or energy content?

Heat entering a sample raises its temperature, by an amount set by the sample's mass and identity. The relation q = m·c·ΔT counts those joules, in these same units.

Dr. Karmach

2 · Exothermic & Endothermic

Classify any process as exothermic or endothermic — from the sign of ΔH, from an energy diagram, or from a heat term written into the equation — and state which way heat flows between system and surroundings.

Dr. Karmach

Two pouches from the drugstore

Snap the pouch inside a hand warmer and it climbs to 54 °C. Snap a cold pack and it drops near freezing. Sealed chemicals drive both changes.

Dr. Karmach

The system and its surroundings

The reaction is the system; the flask, your hand, the room are the surroundings. ΔH records the system's heat: out negative, in positive. The sign follows the system, not your hand.

Dr. Karmach

Enthalpy

Enthalpy, H, is the heat content of a system. A change in it, ΔH, equals the heat of the process at constant pressure. An open flask or a pouch in your hand qualifies.

ΔH = heat of the process at constant pressure
heat out of the system → ΔH negative · heat in → ΔH positive
Dr. Karmach

Exothermic and endothermic

exothermic — heat exits the system
surroundings warm up · ΔH negative · burning fuel, the hand-warmer pouch
endothermic — heat enters the system
surroundings cool down · ΔH positive · melting ice, the cold-pack pouch

Both names describe the system. A process that sends heat out is exothermic; a process that takes heat in is endothermic. The surroundings show the opposite change.

Dr. Karmach

Energy diagrams

An energy diagram plots energy against reaction progress. Products below the reactants: the difference left as heat, exothermic. Products above: the difference came in as heat, endothermic.

Dr. Karmach

Heat written into the equation

CH₄(g) + 2 O₂(g) → CO₂(g) + 2 H₂O(l) + 890 kJ
heat on the product side — it leaves with the products · ΔH = −890 kJ · exothermic
2 H₂O(l) + 572 kJ → 2 H₂(g) + O₂(g)
heat on the reactant side — it must be supplied · ΔH = +572 kJ · endothermic

A thermochemical equation may carry its heat in-line. Product side: heat released, exothermic. Reactant side: heat absorbed, endothermic. The separate ΔH states the same fact with a sign.

Dr. Karmach

The method

  1. Name the system. The process is the system; all else is surroundings.
  2. Find the heat's direction. From the ΔH sign, diagram levels, or heat term.
  3. State the verdict. Heat out: exothermic, ΔH negative. Heat in: endothermic, ΔH positive.
Dr. Karmach

Worked example 1 — a hand warmer's reaction

4 Fe(s) + 3 O₂(g) → 2 Fe₂O₃(s) · ΔH = −1648 kJ
given: the equation and its ΔH · wanted: the verdict and the heat's direction

Inside a hand warmer, iron powder reacts with oxygen from the air.

Classify the reaction and state which way heat flows.

Dr. Karmach

Worked example 1 — solution

4 Fe(s) + 3 O₂(g) → 2 Fe₂O₃(s) · ΔH = −1648 kJ
given: the equation and its ΔH

Step 1 · Name the system

The iron and oxygen are the system. The pouch, the air, your cold hands: surroundings.

Dr. Karmach

Worked example 1 — solution

4 Fe(s) + 3 O₂(g) → 2 Fe₂O₃(s) · ΔH = −1648 kJ
given: the equation and its ΔH
Step 1 · Name the system Step 2 · Find the heat's direction

ΔH is negative: −1648 kJ. Negative marks heat leaving the system.

Dr. Karmach

Worked example 1 — solution

4 Fe(s) + 3 O₂(g) → 2 Fe₂O₃(s) · ΔH = −1648 kJ
given: the equation and its ΔH
Step 1 · Name the system Step 2 · Find the heat's direction Step 3 · State the verdict
4 Fe(s) + 3 O₂(g) → 2 Fe₂O₃(s) + 1648 kJ → exothermic
heat out · ΔH = −1648 kJ · the surroundings — your hands — warm up
Dr. Karmach

Worked example 1 — solution

4 Fe(s) + 3 O₂(g) → 2 Fe₂O₃(s) · ΔH = −1648 kJ
given: the equation and its ΔH
Step 1 · Name the system Step 2 · Find the heat's direction Step 3 · State the verdict
4 Fe(s) + 3 O₂(g) → 2 Fe₂O₃(s) + 1648 kJ → exothermic
heat out · ΔH = −1648 kJ · the surroundings — your hands — warm up
The pouch warms your hand: the surroundings gain exactly the heat the system loses. A negative ΔH and a warming hand agree.
Dr. Karmach

Worked example 2 — hydrogen peroxide decomposes

2 H₂O₂(l) → 2 H₂O(l) + O₂(g)
given: the energy diagram · wanted: the verdict and ΔH with its sign

Hydrogen peroxide fizzing on a cut breaks down into water and oxygen. Classify the process from the diagram and give ΔH.

Dr. Karmach

Worked example 2 — solution

Step 1 · Name the system

The decomposing peroxide is the system; the cut, the skin, the air are surroundings.

Dr. Karmach

Worked example 2 — solution


Step 1 · Name the system
Step 2 · Find the heat's direction

The products sit 196 kJ below the reactants. That difference left the system as heat.

Dr. Karmach

Worked example 2 — solution


Step 1 · Name the system
Step 2 · Find the heat's direction
Step 3 · State the verdict

2 H₂O₂(l) → 2 H₂O(l) + O₂(g) → exothermic
products lower · heat out · ΔH = −196 kJ
Dr. Karmach

Worked example 2 — solution


Step 1 · Name the system
Step 2 · Find the heat's direction
Step 3 · State the verdict

2 H₂O₂(l) → 2 H₂O(l) + O₂(g) → exothermic
products lower · heat out · ΔH = −196 kJ
Downhill on an energy diagram is heat out. The products hold less energy than the reactants, and the fizzing cut warms slightly.
Dr. Karmach

Your turn — photosynthesis

6 CO₂(g) + 6 H₂O(l) + 2803 kJ → C₆H₁₂O₆(s) + 6 O₂(g)
a leaf builds glucose; the 2803 kJ arrives as sunlight
step question answer
1 · name the system what is changing? the CO₂ and water becoming glucose
2 · find the heat's direction which side carries the heat term? the side; heat the system
3 · state the verdict · ΔH = kJ

Complete the three steps.

Dr. Karmach

Your turn — photosynthesis

6 CO₂(g) + 6 H₂O(l) + 2803 kJ → C₆H₁₂O₆(s) + 6 O₂(g)
a leaf builds glucose; the 2803 kJ arrives as sunlight
step question answer
1 · name the system what is changing? the CO₂ and water becoming glucose
2 · find the heat's direction which side carries the heat term? the side; heat the system
3 · state the verdict · ΔH = kJ

Complete the three steps.

photosynthesis → endothermic
heat term on the reactant side · heat enters the system · ΔH = +2803 kJ
Dr. Karmach

Where this goes wrong

Reading the sign from your hand. A cold pack chills your skin, and the chill gets recorded as heat lost: ΔH = −26 kJ. The skin is surroundings. Its loss is the system's gain: ΔH = +26 kJ.
Pairing a label with the opposite flow. "Exothermic, and it absorbs heat" contradicts itself. The label names the flow: exothermic releases heat, endothermic absorbs it. One verdict carries both parts.
Reading the heat term from the wrong side. In 2 H₂O(l) + 572 kJ → 2 H₂(g) + O₂(g), the 572 kJ gets reported as released. It sits with the reactants, so it is consumed: absorbed, ΔH = +572 kJ.
Calling the higher level the bigger release. Height on an energy diagram is energy stored, not energy given off. Products above the reactants means the system took energy in: endothermic, ΔH positive.
Dr. Karmach

Practice 1

2 SO₂(g) + O₂(g) → 2 SO₃(g) · ΔH = −198 kJ
given: the equation and its ΔH

Sulfur dioxide converts to sulfur trioxide during sulfuric acid manufacture. Which statement describes the reaction?

  1. Endothermic — heat is absorbed by the system from the surroundings
  2. Exothermic — heat is absorbed by the system from the surroundings
  3. Exothermic — heat is released by the system to the surroundings
  4. Endothermic — heat is released by the system to the surroundings
Dr. Karmach

Practice 1 — answer: C

ΔH = −198 kJ → negative → heat out → exothermic — answer C
heat released by the system · the surroundings warm up

A flipped the sign convention: heat absorbed would be counted into the system, +198 kJ, not −198 kJ. B paired the right label with the wrong flow: exothermic means heat exits the system. D paired the right flow with the wrong label: a heat-releasing reaction is exothermic.

Negative ΔH, heat out, exothermic, warmer surroundings: four readings of the same event.
Dr. Karmach

Worked example 3 — the cold pack

NH₄NO₃(s) → NH₄NO₃(aq)
given: the pouch turns icy in your hand · wanted: the verdict and the sign of ΔH

Snapping the pack lets ammonium nitrate dissolve in water, and the pouch turns icy.

A common first answer: the pack is cold, so it is losing heat — exothermic. Test it.

Dr. Karmach

Worked example 3 — solution

NH₄NO₃(s) → NH₄NO₃(aq)
the pouch turns icy in your hand

A common first answer

cold pack, so the pack is losing heat → exothermic?
cold marks heat leaving the pack only if the pack were the surroundings ✗

The cold skin is the evidence. Your hand is losing heat, and the hand is surroundings, not system.

Dr. Karmach

Worked example 3 — solution

NH₄NO₃(s) → NH₄NO₃(aq)
the pouch turns icy in your hand
A common first answer
cold pack, so the pack is losing heat → exothermic?
cold marks heat leaving the pack only if the pack were the surroundings ✗
Step 1 · Name the system

The dissolving salt and water are the system. The pouch, your hand: surroundings.

Dr. Karmach

Worked example 3 — solution

NH₄NO₃(s) → NH₄NO₃(aq)
the pouch turns icy in your hand
A common first answer
cold pack, so the pack is losing heat → exothermic?
cold marks heat leaving the pack only if the pack were the surroundings ✗
Step 1 · Name the system Step 2 · Find the heat's direction

Your hand cools: heat is leaving the surroundings and entering the system.

Dr. Karmach

Worked example 3 — solution

NH₄NO₃(s) → NH₄NO₃(aq)
the pouch turns icy in your hand
A common first answer
cold pack, so the pack is losing heat → exothermic?
cold marks heat leaving the pack only if the pack were the surroundings ✗
Step 1 · Name the system Step 2 · Find the heat's direction Step 3 · State the verdict
NH₄NO₃(s) → NH₄NO₃(aq) → endothermic
heat in · measured ΔH = +26 kJ per mole dissolved · the surroundings — your hand — cool
Dr. Karmach

Worked example 3 — solution

NH₄NO₃(s) → NH₄NO₃(aq)
the pouch turns icy in your hand
A common first answer
cold pack, so the pack is losing heat → exothermic?
cold marks heat leaving the pack only if the pack were the surroundings ✗
Step 1 · Name the system Step 2 · Find the heat's direction Step 3 · State the verdict
NH₄NO₃(s) → NH₄NO₃(aq) → endothermic
heat in · measured ΔH = +26 kJ per mole dissolved · the surroundings — your hand — cool
The pack feels cold *because* it absorbs heat. A cooling hand is the surroundings' loss and the system's gain: ΔH = +26 kJ, never −26 kJ.
Dr. Karmach

Take-home: your hand is the surroundings

feels hot → the surroundings are gaining heat → the system is losing it
exothermic · ΔH negative
feels cold → the surroundings are losing heat → the system is gaining it
endothermic · ΔH positive

Skin and thermometers sit in the surroundings. They report the surroundings' change, and the system did the opposite. Feels cold: the system is absorbing heat — endothermic, ΔH positive.

Dr. Karmach

Practice 2

NH₄Cl(s) → NH₄Cl(aq)
given: the beaker turns cold as the solid dissolves · wanted: the verdict and the sign of ΔH

Ammonium chloride is stirred into room-temperature water, and the beaker turns cold to the touch. Which statement describes the dissolving process?

  1. Exothermic, ΔH negative — the beaker is losing heat, so the process releases heat
  2. Endothermic, ΔH negative — the process absorbs heat, and absorbed heat is written with a minus sign
  3. Exothermic, ΔH positive — the temperature change proves heat was produced
  4. Endothermic, ΔH positive — the dissolving salt pulls heat in from the beaker, the bench, and your hand
Dr. Karmach

Practice 2 — answer: D

cold beaker → surroundings losing heat → heat entering the system → endothermic, ΔH = +15 kJ — answer D
measured: dissolving one mole of NH₄Cl absorbs 15 kJ

A read the sign from the cold feeling: the beaker and your hand are surroundings, and their loss is the system's gain — +15 kJ, not −15 kJ. B matched the label but not the sign: absorbed heat counts into the system as positive. C matched the sign but not the label: a heat-absorbing process is endothermic.

Cold to the touch means your hand is donating heat. A process that takes heat in is endothermic, whatever your skin reports.
Dr. Karmach

Check yourself

  1. Propane burns in a camp stove: C₃H₈(g) + 5 O₂(g) → 3 CO₂(g) + 4 H₂O(l), ΔH = −2220 kJ. State the verdict and which way heat flows.
  2. An energy diagram shows the products 178 kJ above the reactants. Give the sign of ΔH and the verdict.

Exothermic or endothermic names the heat's direction. How much heat a sample gains or loses depends on its mass, its substance, and its temperature change: q = m·c·ΔT.

Dr. Karmach

3 · Specific Heat & q = mcΔT

Use q = m·c·ΔT to find the heat, the mass, the specific heat, or the temperature change, with ΔT measured final minus initial so the sign of q shows which way the heat flowed.

Dr. Karmach

One burner, two temperatures

Five minutes in, the iron handle is too hot to touch; the water is barely warm. Iron needs far less heat than water for each degree it climbs.

Dr. Karmach

Specific heat: joules per gram per degree

specific heat c — the heat that raises 1 g of a substance by 1 °C
J/g·°C — water 4.184 · ethyl alcohol 2.138 · aluminum 0.897 · iron 0.449 · copper 0.385 · gold 0.129 · lead 0.128

Heat flowing in raises a substance's temperature; heat flowing out lowers it. The joules needed to move each gram by one degree are fixed for each substance: its specific heat, c.

Dr. Karmach

The heat equation

Three factors set the heat: the mass m, the substance's specific heat c, and the temperature change ΔT. One equation, four solvable quantities.

Dr. Karmach

ΔT carries a sign

ΔT = Tfinal − Tinitial
heating 20.0 → 50.0 °C: ΔT = +30.0 °C · cooling 50.0 → 20.0 °C: ΔT = −30.0 °C

ΔT is final minus initial, in that order. A cooling sample has a negative ΔT, so q comes out negative: the sample released heat. The sign records the direction of the flow.

Dr. Karmach

Specific heat c vs heat capacity C

C = m · c
a 250-g water sample: C = 250 g × 4.184 J/g·°C = 1046 J/°C · c stays 4.184 J/g·°C for any amount of water

Specific heat describes each gram of a material. Heat capacity C describes one whole object: the joules that raise that object, all of it, by 1 °C.

Dr. Karmach

The method

  1. List the pieces: m, c, ΔT = Tfinal − Tinitial. Mark the unknown.
  2. Rearrange for the unknown before numbers go in.
  3. Substitute and cancel units.
  4. Check the sign: cooling means negative ΔT and negative q.
Dr. Karmach

Worked example 1 — heat to warm water

q = m · c · ΔT
given: 250 g water · c = 4.184 J/g·°C · 20.0 °C → 50.0 °C · wanted: q

A kettle warms 250 g of water from 20.0 °C to 50.0 °C. How much heat does the water absorb? (c of water: 4.184 J/g·°C)

List the pieces: m, c, and ΔT = Tfinal − Tinitial.

Dr. Karmach

Worked example 1 — solution

q = m · c · ΔT
given: 250 g water · c = 4.184 J/g·°C · 20.0 °C → 50.0 °C · wanted: q

Step 1 · List the pieces

m = 250 g. c = 4.184 J/g·°C. ΔT = 50.0 − 20.0 = +30.0 °C. The unknown is q.

Dr. Karmach

Worked example 1 — solution

q = m · c · ΔT
given: 250 g water · c = 4.184 J/g·°C · 20.0 °C → 50.0 °C · wanted: q
Step 1 · List the pieces Step 2 · Rearrange for the unknown

q already stands alone on its side of the equation; no rearranging is needed.

Dr. Karmach

Worked example 1 — solution

q = m · c · ΔT
given: 250 g water · c = 4.184 J/g·°C · 20.0 °C → 50.0 °C · wanted: q
Step 1 · List the pieces Step 2 · Rearrange for the unknown Step 3 · Substitute and cancel units
q = 250 g × 4.184 J1 g·°C × 30.0 °C = 31,400 J
Dr. Karmach

Worked example 1 — solution

q = m · c · ΔT
given: 250 g water · c = 4.184 J/g·°C · 20.0 °C → 50.0 °C · wanted: q
Step 1 · List the pieces Step 2 · Rearrange for the unknown Step 3 · Substitute and cancel units
q = 250 g × 4.184 J1 g·°C × 30.0 °C = 31,400 J
Step 4 · Check the sign
The water warmed, so ΔT and q are both positive: 31,400 J (31.4 kJ) absorbed. More grams or more degrees would cost more heat. ✓
Dr. Karmach

Worked example 2 — find c, identify the substance

q = m · c · ΔT
given: q = 1347 J · m = 125 g · 22.0 °C → 46.0 °C · wanted: c

A 125-g metal block absorbs 1347 J as it warms from 22.0 °C to 46.0 °C. Candidate specific heats, in J/g·°C:

aluminum iron copper gold lead
0.897 0.449 0.385 0.129 0.128

Find the block's specific heat and match it to the table.

Dr. Karmach

Worked example 2 — solution

q = m · c · ΔT
given: q = 1347 J · m = 125 g · 22.0 °C → 46.0 °C · wanted: c

Step 1 · List the pieces

q = 1347 J. m = 125 g. ΔT = 46.0 − 22.0 = +24.0 °C. The unknown is c.

Dr. Karmach

Worked example 2 — solution

q = m · c · ΔT
given: q = 1347 J · m = 125 g · 22.0 °C → 46.0 °C · wanted: c
Step 1 · List the pieces Step 2 · Rearrange for the unknown
q = m · c · ΔT → c = qm · ΔT

Divide both sides by m·ΔT before any numbers go in.

Dr. Karmach

Worked example 2 — solution

q = m · c · ΔT
given: q = 1347 J · m = 125 g · 22.0 °C → 46.0 °C · wanted: c
Step 1 · List the pieces Step 2 · Rearrange for the unknown
q = m · c · ΔT → c = qm · ΔT
Step 3 · Substitute and cancel units
c = 1347 J125 g × 24.0 °C = 0.449 J/g·°C

No unit cancels here; they assemble into J/g·°C — the unit of a specific heat.

Dr. Karmach

Worked example 2 — solution

q = m · c · ΔT
given: q = 1347 J · m = 125 g · 22.0 °C → 46.0 °C · wanted: c
Step 1 · List the pieces Step 2 · Rearrange for the unknown
q = m · c · ΔT → c = qm · ΔT
Step 3 · Substitute and cancel units
c = 1347 J125 g × 24.0 °C = 0.449 J/g·°C
0.449 J/g·°C sits in the range of a metal's specific heat — ready to identify. ✓
Dr. Karmach

Worked example 2 — identify the metal

c = 0.449 J/g·°C
from q = 1347 J · m = 125 g · ΔT = +24.0 °C, both positive

Match the property

aluminum iron copper gold lead
0.897 0.449 0.385 0.129 0.128

Only iron matches 0.449 J/g·°C. The block is iron.

Dr. Karmach

Worked example 2 — identify the metal

c = 0.449 J/g·°C
from q = 1347 J · m = 125 g · ΔT = +24.0 °C, both positive
Match the property
aluminum iron copper gold lead
0.897 0.449 0.385 0.129 0.128

Step 4 · Check the sign

The block warmed: ΔT and q are both positive. Plug back in: 125 g × 0.449 J/g·°C × 24.0 °C returns 1347 J. ✓
Dr. Karmach

Your turn — mass of ethyl alcohol

q = m · c · ΔT
given: q = 8020 J · c = 2.138 J/g·°C · 18.0 °C → 43.0 °C · wanted: m

A sample of ethyl alcohol (c = 2.138 J/g·°C) absorbs 8020 J and warms from 18.0 °C to 43.0 °C.

m = qc · ΔT = 8020 J J/g·°C × °C = g

Fill in c and ΔT = Tfinal − Tinitial, then compute the mass.

Dr. Karmach

Your turn — mass of ethyl alcohol

q = m · c · ΔT
given: q = 8020 J · c = 2.138 J/g·°C · 18.0 °C → 43.0 °C · wanted: m

A sample of ethyl alcohol (c = 2.138 J/g·°C) absorbs 8020 J and warms from 18.0 °C to 43.0 °C.

m = qc · ΔT = 8020 J J/g·°C × °C = g

Fill in c and ΔT = Tfinal − Tinitial, then compute the mass.

m = 8020 J2.138 J/g·°C × 25.0 °C = 150. g
2.138 J warms one gram by one degree, so 8020 J spread over 25.0 degrees warms about 150 g. ✓
Dr. Karmach

Where this goes wrong

q = m · c · ΔT
250 g water · c = 4.184 J/g·°C · 20.0 → 50.0 °C · correct q = 31,400 J
Leaving out the mass. 4.184 × 30.0 = 126 J is the heat for a single gram. The sample has 250 of them. All three factors multiply: q = m·c·ΔT = 31,400 J.
Leaving out the temperature change. 250 × 4.184 = 1046 J warms the water by one degree only. Multiply by the full ΔT of 30.0 °C.
Dividing by the specific heat. 250 × 30.0 ÷ 4.184 = 1790, and its units are g²·°C²/J, not joules. c multiplies on top: (4.184 J / 1 g·°C).
Subtracting the temperatures in the wrong order. ΔT = 20.0 − 50.0 = −30.0 °C gives q = −31,400 J: heat released by water that is warming. ΔT is Tfinal − Tinitial.
Dr. Karmach

Practice 1

q = m · c · ΔT
given: 75.0 g copper · c = 0.385 J/g·°C · ΔT = +20.0 °C · wanted: q

A 75.0-g copper fitting warms by 20.0 °C as hot water flows past it. How much heat does the copper absorb? (c of copper: 0.385 J/g·°C)

  1. 7.70 J
  2. 28.9 J
  3. 578 J
  4. 3900 J
Dr. Karmach

Practice 1 — answer: C

q = m · c · ΔT
given: 75.0 g copper · c = 0.385 J/g·°C · ΔT = +20.0 °C
q = 75.0 g × 0.385 J1 g·°C × 20.0 °C = 578 J — answer C

A left out the mass: 0.385 × 20.0 = 7.70 J warms one gram. B left out the temperature change: 75.0 × 0.385 = 28.9 J is one degree's worth. D divided by the specific heat: 75.0 × 20.0 ÷ 0.385 = 3900, with units g²·°C²/J.

Copper takes only 0.385 J per gram per degree, but 75 grams and 20 degrees multiply that into hundreds of joules. ✓
Dr. Karmach

Worked example 3 — final temperature of a cooling sample

q = m · c · ΔT
given: 150 g aluminum · c = 0.897 J/g·°C · Tinitial = 95.0 °C · releases 8100 J · wanted: Tfinal

A 150-g aluminum pan lid at 95.0 °C releases 8100 J as it cools. What is its final temperature? (c of aluminum: 0.897 J/g·°C)

A common first attempt: substitute 8100 J with no sign. Test the result.

Dr. Karmach

Worked example 3 — solution

q = m · c · ΔT
given: 150 g aluminum · c = 0.897 J/g·°C · Tinitial = 95.0 °C · releases 8100 J · wanted: Tfinal

A common first attempt

ΔT = +8100 J150 g × 0.897 J/g·°C = +60.2 °C → Tfinal = 95.0 + 60.2 = 155.2 °C ✗

A lid that is releasing heat cannot end up hotter. The sign of q was dropped.

Dr. Karmach

Worked example 3 — solution

q = m · c · ΔT
given: 150 g aluminum · c = 0.897 J/g·°C · Tinitial = 95.0 °C · releases 8100 J · wanted: Tfinal
A common first attempt
ΔT = +8100 J150 g × 0.897 J/g·°C = +60.2 °C → Tfinal = 95.0 + 60.2 = 155.2 °C ✗
Step 1 · List the pieces

Released heat leaves the sample, so q = −8100 J. m = 150 g. c = 0.897 J/g·°C. The unknown is ΔT, then Tfinal.

Dr. Karmach

Worked example 3 — solution

q = m · c · ΔT
given: 150 g aluminum · c = 0.897 J/g·°C · Tinitial = 95.0 °C · releases 8100 J · wanted: Tfinal
A common first attempt
ΔT = +8100 J150 g × 0.897 J/g·°C = +60.2 °C → Tfinal = 95.0 + 60.2 = 155.2 °C ✗
Step 1 · List the pieces Step 2 · Rearrange for the unknown
ΔT = qm · c , then Tfinal = Tinitial + ΔT
With q entered as −8100 J, the formula is set to return a negative ΔT: a temperature drop. ✓
Dr. Karmach

Worked example 3 — final temperature

q = −8100 J released · 150 g aluminum · c = 0.897 J/g·°C
Tinitial = 95.0 °C · wanted: Tfinal

Step 3 · Substitute and cancel units

ΔT = −8100 J150 g × 0.897 J/g·°C = −8100 J134.55 J/°C = −60.2 °C
Tfinal = 95.0 °C + (−60.2 °C) = 34.8 °C
Dr. Karmach

Worked example 3 — final temperature

q = −8100 J released · 150 g aluminum · c = 0.897 J/g·°C
Tinitial = 95.0 °C · wanted: Tfinal
Step 3 · Substitute and cancel units
ΔT = −8100 J150 g × 0.897 J/g·°C = −8100 J134.55 J/°C = −60.2 °C
Tfinal = 95.0 °C + (−60.2 °C) = 34.8 °C
Step 4 · Check the sign
Released heat means negative q, negative ΔT, and a lower final temperature: 95.0 → 34.8 °C. ✓ The signless route predicted 155.2 °C, a cooling lid ending hotter. ✗
Dr. Karmach

Take-home: ΔT is final minus initial

warming: 20.0 °C → 50.0 °C · ΔT = 50.0 − 20.0 = +30.0 °C · q positive
heat absorbed ✓
cooling: 95.0 °C → 34.8 °C · ΔT = 34.8 − 95.0 = −60.2 °C · q negative
heat released ✓

ΔT is always Tfinal − Tinitial, and released heat enters as negative q. The sign is part of the quantity; it records which way the heat flowed.

Dr. Karmach

Practice 2

q = m · c · ΔT
given: 100. g lead · c = 0.128 J/g·°C · 62.0 °C → 22.0 °C · wanted: q

A 100.-g lead sinker at 62.0 °C drops into a stream and cools to 22.0 °C. What is q for the lead? (c of lead: 0.128 J/g·°C)

  1. 512 J
  2. -512 J
  3. 12.8 J
  4. -5.12 J
Dr. Karmach

Practice 2 — answer: B

q = m · c · ΔT
given: 100. g lead · c = 0.128 J/g·°C · ΔT = 22.0 − 62.0 = −40.0 °C
q = 100. g × 0.128 J1 g·°C × (−40.0 °C) = −512 J — answer B

A subtracted the temperatures in the wrong order: 62.0 − 22.0 = +40.0 °C gives +512 J, heat absorbed by a cooling sinker. C stopped at m × c: 100. × 0.128 = 12.8 J, one degree's worth with no sign. D left out the mass: 0.128 × (−40.0) = −5.12 J, the heat for a single gram.

The sinker cooled 40 degrees, so it released heat: q must be negative. ✓
Dr. Karmach

Check yourself

  1. Water cools from 50.0 °C to 20.0 °C. Write ΔT with its sign. What is the sign of q, and what does it say about the heat?
  2. Solve q = m·c·ΔT for m, in symbols. Which units cancel, and which unit survives?

Two samples can receive the same heat and change temperature by different amounts. Equal masses, equal q: the substance with the smaller specific heat shows the larger ΔT.

Dr. Karmach

4 · Comparing Specific Heats

For the same heat and mass, read specific heat in reverse — the smaller c, the larger the temperature change — reasoning from a c table where a comparison settles it and confirming with ΔT = q/mc when a number is wanted.

Dr. Karmach

The beach, twelve hours apart

At noon the sand scorches bare feet while the ocean stays cool. By midnight the sand is cold, and the water is the warm place to be.

Dr. Karmach

Same heat and mass: c and ΔT trade off

Give equal masses the same heat. Their temperature changes are not equal. Specific heat sits in the denominator of ΔT = q/mc, so the smaller c, the larger the temperature change.

ΔT = q / (m · c)
same q, same m: c in the denominator — smaller c, larger ΔT
Dr. Karmach

Ranking substances by specific heat

Every substance has its own specific heat. Water's is several times any metal's. Read the table in reverse: the higher the specific heat, the smaller the temperature change from the same heat.

water 4.184 · ethyl alcohol 2.138 · aluminum 0.897 · iron 0.449 · copper 0.385 · silver 0.235 · gold 0.129 · lead 0.128
specific heat c, in J/g·°C — smaller c, larger temperature change for the same heat and mass
Dr. Karmach

Why water resists temperature swings

Water's high specific heat means it soaks up heat with only a small temperature rise, and gives it back slowly. Coastlines stay mild through the day. Engines and reactors use water to carry heat away.

2092 J into 100 g of each: water rises +5.0 °C · iron rises +46.6 °C
water's specific heat is 9.3× iron's, so the same heat moves it 9.3× less
Dr. Karmach

The method

  1. Compare the specific heats. Same heat and mass: smaller c means larger ΔT.
  2. Name the response. The substance with the smaller c swings more.
  3. Confirm with ΔT = q/mc. c and m are in the denominator.
Dr. Karmach

Worked example 1 — two metals, one burner

ΔT = q / (m · c)
100 g each of lead and aluminum · q = 900 J each · c: lead 0.128, aluminum 0.897 J/g·°C · wanted: which ends hotter

Two blocks, the same 100 g and the same starting temperature, each absorb 900 J.

Compare the specific heats, then confirm both temperature changes.

Dr. Karmach

Worked example 1 — solution

ΔT = q / (m · c)
100 g each · q = 900 J each · c: lead 0.128, aluminum 0.897 J/g·°C

Step 1 · Compare the specific heats

Lead's specific heat, 0.128, is far smaller than aluminum's, 0.897.

Dr. Karmach

Worked example 1 — solution

ΔT = q / (m · c)
100 g each · q = 900 J each · c: lead 0.128, aluminum 0.897 J/g·°C
Step 1 · Compare the specific heats Step 2 · Name the response

The same heat spread over the smaller cost per degree gives the bigger rise. Lead swings more.

Dr. Karmach

Worked example 1 — solution

ΔT = q / (m · c)
100 g each · q = 900 J each · c: lead 0.128, aluminum 0.897 J/g·°C
Step 1 · Compare the specific heats Step 2 · Name the response Step 3 · Confirm with ΔT = q/mc
lead: ΔT = 900 J100 g × 0.128 J/g·°C = 900 J12.8 J/°C = +70.3 °C
aluminum: ΔT = 900 J100 g × 0.897 J/g·°C = 900 J89.7 J/°C = +10.0 °C
Dr. Karmach

Worked example 1 — solution

ΔT = q / (m · c)
100 g each · q = 900 J each · c: lead 0.128, aluminum 0.897 J/g·°C
Step 1 · Compare the specific heats Step 2 · Name the response Step 3 · Confirm with ΔT = q/mc
lead: ΔT = 900 J100 g × 0.128 J/g·°C = 900 J12.8 J/°C = +70.3 °C
aluminum: ΔT = 900 J100 g × 0.897 J/g·°C = 900 J89.7 J/°C = +10.0 °C
Same mass, same heat, and lead climbs 70.3 °C while aluminum climbs 10.0 °C. Smaller c, bigger swing. ✓
Dr. Karmach

Worked example 2 — rank three by final temperature

ΔT = q / (m · c)
50.0 g each of water, aluminum, copper · q = 1000 J each · start 20.0 °C · c: water 4.184, aluminum 0.897, copper 0.385 J/g·°C

Three samples, the same 50.0 g, all starting at 20.0 °C, each absorb 1000 J.

A common first answer: water has the highest specific heat, so water climbs the most. Rank the three final temperatures.

Dr. Karmach

Worked example 2 — solution

ΔT = q / (m · c)
50.0 g each · 1000 J each · start 20.0 °C

Step 1 · Compare the specific heats

c sits in the denominator, so the highest specific heat gives the smallest ΔT. Water, the highest, climbs least — the first answer had it reversed.

Dr. Karmach

Worked example 2 — solution

ΔT = q / (m · c)
50.0 g each · 1000 J each · start 20.0 °C
Step 1 · Compare the specific heats Step 2 · Name the response
Dr. Karmach

Worked example 2 — solution

ΔT = q / (m · c)
50.0 g each · 1000 J each · start 20.0 °C
Step 1 · Compare the specific heats Step 2 · Name the response
Copper has the smallest c, so copper swings most; aluminum is in between; water swings least. ✓
Dr. Karmach

Worked example 2 — confirm with ΔT = q/mc

Step 3 · Confirm with ΔT = q/mc

water: ΔT = 1000 J50.0 g × 4.184 J/g·°C = +4.78 °C → 24.8 °C
aluminum: ΔT = 1000 J50.0 g × 0.897 J/g·°C = +22.3 °C → 42.3 °C
copper: ΔT = 1000 J50.0 g × 0.385 J/g·°C = +51.9 °C → 72.0 °C
Dr. Karmach

Worked example 2 — confirm with ΔT = q/mc

Step 3 · Confirm with ΔT = q/mc

water: ΔT = 1000 J50.0 g × 4.184 J/g·°C = +4.78 °C → 24.8 °C
aluminum: ΔT = 1000 J50.0 g × 0.897 J/g·°C = +22.3 °C → 42.3 °C
copper: ΔT = 1000 J50.0 g × 0.385 J/g·°C = +51.9 °C → 72.0 °C
Lowest specific heat ends hottest: copper 72.0 > aluminum 42.3 > water 24.8 °C. ✓
Dr. Karmach

Take-home: high c, smaller swing

50.0 g each, 1000 J each: water ΔT = +4.78 °C · copper ΔT = +51.9 °C
water's c is the highest, so its swing is the smallest — the reverse of "heats fastest"

"Higher specific heat, faster heating" gets it backwards. A high specific heat means more joules for every degree. For the same heat and mass, the high-c substance changes temperature the least.

Dr. Karmach

Your turn — gold and silver

Equal 40.0-g samples of gold and silver each absorb 250 J. Fill each c into ΔT = q/(m·c), then compute both temperature changes.

gold: ΔT = 250 J40.0 g × J/g·°C = °C
silver: ΔT = 250 J40.0 g × J/g·°C = °C
Dr. Karmach

Your turn — gold and silver

gold: ΔT = 250 J40.0 g × J/g·°C = °C
silver: ΔT = 250 J40.0 g × J/g·°C = °C
gold: ΔT = 250 J40.0 g × 0.129 J/g·°C = +48.4 °C
silver: ΔT = 250 J40.0 g × 0.235 J/g·°C = +26.6 °C
Gold's smaller c gives the larger swing: 48.4 °C against silver's 26.6 °C. ✓
Dr. Karmach

Where this goes wrong

Equal heat, equal temperature. Equal q into equal mass does not give equal ΔT. 1000 J into 50.0 g of water raises it 4.78 °C; the same 1000 J into 50.0 g of copper raises it 51.9 °C. The specific heats differ, so the temperature changes differ.
Right substance, backwards reason. Naming the low-c substance as the one that ends hotter "because it stores more heat per gram." It stores less per gram — lead takes 0.128 J to warm a gram by a degree, aluminum 0.897 J. That low cost per degree is exactly why lead swings more.
Comparing c when the masses differ. The rule "smaller c wins" assumes equal mass. When the masses are not equal, mass is in the denominator too. Compare the whole sample's m·c, or run ΔT = q/mc for each.
Dr. Karmach

Practice 1

equal masses · same heat absorbed · same start
iron c = 0.449 J/g·°C · ethyl alcohol c = 2.138 J/g·°C

Equal masses of iron and ethyl alcohol start at the same temperature and each absorbs the same heat. Which ends up at the higher final temperature?

  1. Ethyl alcohol — its larger specific heat means its temperature climbs faster.
  2. Iron — it stores more heat per gram, so it ends up hotter.
  3. Iron — its smaller specific heat means the same heat produces a larger temperature rise.
  4. Both reach the same final temperature, since they absorbed equal heat.
Dr. Karmach

Practice 1 — answer: C

ΔT = q / (m · c) — same q, same m
iron c = 0.449 → smaller denominator → larger ΔT · ethyl alcohol c = 2.138 → larger denominator → smaller ΔT

Iron has the smaller specific heat, so the same heat produces the larger temperature rise — answer C.

A inverted the reading: a larger specific heat means more joules per degree, so ethyl alcohol climbs less, not faster. B named the right substance for the wrong reason — iron stores less heat per gram, and that low cost per degree is why it warms more. D forgets that equal heat into equal but different substances gives different ΔT, because the two c values differ.

Smaller c, bigger swing: iron ends hotter. ✓
Dr. Karmach

Worked example 3 — when the masses differ

ΔT = q / (m · c)
200 g copper (c = 0.385) and 50.0 g aluminum (c = 0.897) · q = 2000 J each · wanted: which ends hotter

A 200 g copper block and a 50.0 g aluminum block each absorb 2000 J.

A common first answer: copper has the smaller specific heat, so copper wins. Check it — the masses are not equal.

Dr. Karmach

Worked example 3 — solution

ΔT = q / (m · c)
200 g copper (c = 0.385) · 50.0 g aluminum (c = 0.897) · q = 2000 J each

Step 1 · Compare the specific heats

Copper's c, 0.385, is smaller than aluminum's, 0.897. But mass is in the denominator too, and the copper sample is four times heavier.

Dr. Karmach

Worked example 3 — solution

ΔT = q / (m · c)
200 g copper (c = 0.385) · 50.0 g aluminum (c = 0.897) · q = 2000 J each
Step 1 · Compare the specific heats Step 2 · Name the response

Compare the whole sample's m·c: copper 200 × 0.385 = 77.0 J/°C, aluminum 50.0 × 0.897 = 44.85 J/°C. Aluminum's is smaller, so aluminum swings more.

Dr. Karmach

Worked example 3 — solution

ΔT = q / (m · c)
200 g copper (c = 0.385) · 50.0 g aluminum (c = 0.897) · q = 2000 J each
Step 1 · Compare the specific heats Step 2 · Name the response Step 3 · Confirm with ΔT = q/mc
copper: ΔT = 2000 J200 g × 0.385 J/g·°C = 2000 J77.0 J/°C = +26.0 °C
aluminum: ΔT = 2000 J50.0 g × 0.897 J/g·°C = 2000 J44.85 J/°C = +44.6 °C
Dr. Karmach

Worked example 3 — solution

ΔT = q / (m · c)
200 g copper (c = 0.385) · 50.0 g aluminum (c = 0.897) · q = 2000 J each
Step 1 · Compare the specific heats Step 2 · Name the response Step 3 · Confirm with ΔT = q/mc
copper: ΔT = 2000 J200 g × 0.385 J/g·°C = 2000 J77.0 J/°C = +26.0 °C
aluminum: ΔT = 2000 J50.0 g × 0.897 J/g·°C = 2000 J44.85 J/°C = +44.6 °C
Aluminum ends hotter, 44.6 °C against 26.0 °C, even though copper has the smaller specific heat. When the masses differ, compare m·c, not c alone. ✓
Dr. Karmach

Practice 2

ΔT = q / (m · c)
100 g and 250 g of iron · c = 0.449 J/g·°C · q = 2500 J each · start 20.0 °C

A 100 g iron bar and a 250 g iron bar each absorb 2500 J, both starting at 20.0 °C. What is the final temperature of the 100 g bar?

  1. 75.7 °C
  2. 55.7 °C
  3. 42.3 °C
  4. 22.3 °C
Dr. Karmach

Practice 2 — answer: A

ΔT = q / (m · c)
100 g iron · c = 0.449 J/g·°C · q = 2500 J · start 20.0 °C
ΔT = 2500 J100 g × 0.449 J/g·°C = +55.7 °C → 20.0 + 55.7 = 75.7 °C — answer A

B stopped at the temperature change and forgot to add the 20.0 °C start: 55.7 °C. C used the 250 g mass: 2500 ÷ (250 × 0.449) = 22.3 °C, then 20.0 + 22.3 = 42.3 °C — the heavier bar's answer. D used the 250 g mass and also dropped the start: 22.3 °C.

The lighter bar has the smaller m·c, so the same heat gives it the bigger swing and the higher final temperature. ✓
Dr. Karmach

Check yourself

  1. Equal masses of copper and water absorb the same heat from the same start. Which ends hotter, and which term in ΔT = q/mc decides it?
  2. Two iron bars, 50 g and 150 g, absorb the same heat. Which shows the larger temperature change, and why?

Drop a hot metal into cool water and heat leaves the metal and enters the water until both settle at one temperature. The heat lost by the metal equals the heat gained by the water: calorimetry.

Dr. Karmach

5 · Calorimetry

Use an insulated cup to make energy conservation visible — the heat one body loses equals the heat another gains — and solve for a final temperature or a reaction's enthalpy, with the answer's final temperature landing between the two starting temperatures.

Dr. Karmach

How a Calorie gets measured

A food label's Calorie number is not estimated. The food is burned in a sealed chamber, water around it absorbs the heat, and the temperature rise gives the count.

Dr. Karmach

Heat lost equals heat gained

Nest two foam cups, add a lid and a thermometer, and almost no heat escapes. Whatever heat the hot object loses, the water gains. The two settle at one shared temperature.

Dr. Karmach

Energy is conserved in the cup

Inside the insulated cup, the energy that leaves one body enters the other. Energy is conserved: none is created or destroyed.

qlost + qgained = 0 → heat lost = heat gained
hot body: m·c·(Thigh − Tf) · cold body: m·c·(Tf − Tlow)
Dr. Karmach

The final temperature sits between

Both bodies end at the same temperature. It lands between the two starting temperatures, pulled toward whichever body carries the larger m·c.

water's m·c is usually the largest term
a hot metal in cool water settles close to the water's starting temperature
Dr. Karmach

The method

  1. Heat lost = heat gained. Write m·c·(Thigh − Tf) = m·c·(Tf − Tlow).
  2. Solve for the unknown, usually Tf.
  3. Check Tf lands between the two starting temperatures.
Dr. Karmach

Worked example 1 — mixing hot and cold water

heat lost = heat gained
given: 150 g water at 70 °C · 100 g water at 20 °C · wanted: Tf

Pour 150 g of water at 70 °C into 100 g at 20 °C. Both are water, so both specific heats are 4.184. Find the final temperature.

Dr. Karmach

Worked example 1 — solution

heat lost = heat gained
150 g water at 70 °C · 100 g water at 20 °C

Step 1 · Heat lost = heat gained

The same c on both sides cancels: 150 · (70 − Tf) = 100 · (Tf − 20).

Dr. Karmach

Worked example 1 — solution

heat lost = heat gained
150 g water at 70 °C · 100 g water at 20 °C
Step 1 · Heat lost = heat gained Step 2 · Solve for the unknown
Tf = 150 × 70 + 100 × 20150 + 100 = 12,500250 = 50.0 °C
50.0 °C lands between 20 and 70, and closer to 70 because the hotter sample is heavier. ✓
Dr. Karmach

Worked example 2 — a hot bolt in water

heat lost = heat gained
given: 100 g iron at 90 °C · 150 g water at 20 °C · c: iron 0.449, water 4.184 · wanted: Tf

A 100 g iron bolt at 90 °C drops into 150 g of water at 20 °C.

A common first attempt: add the two heats. Test the result against the thermometer.

Dr. Karmach

Worked example 2 — solution

heat lost = heat gained
100 g iron at 90 °C · 150 g water at 20 °C

A common first attempt

adding the heats: Tf = 14.6 °C ✗
14.6 °C is below both 20 °C and 90 °C — no shared temperature sits outside the pair

Same signs on both heats is the slip. One body loses, the other gains.

Dr. Karmach

Worked example 2 — solution

heat lost = heat gained
100 g iron at 90 °C · 150 g water at 20 °C
A common first attempt
adding the heats: Tf = 14.6 °C ✗
14.6 °C is below both 20 °C and 90 °C — no shared temperature sits outside the pair
Step 1 · Heat lost = heat gained
100 × 0.449 × (90 − Tf) = 150 × 4.184 × (Tf − 20)
The two m·c terms are set: iron's 44.9 against the water's 627.6. Solve for Tf. ✓
Dr. Karmach

Worked example 2 — the final temperature

heat lost = heat gained
m·c: iron 44.9, water 627.6 · start 90 °C and 20 °C

Step 2 · Solve for the unknown

Tf = 44.9 × 90 + 627.6 × 2044.9 + 627.6 = 24.7 °C
24.7 °C sits between 20 and 90, close to the water: its m·c of 627.6 dwarfs the iron's 44.9. ✓
Dr. Karmach

Take-home: one loses, the other gains

heat lost by the hot body = heat gained by the cold body
Tf always lands between the two starting temperatures

Never add the two heats. One body releases, the other absorbs. Set the loss equal to the gain, and a final temperature outside the starting pair means a dropped sign.

Dr. Karmach

Your turn — aluminum into water

Drop 80 g aluminum (c = 0.897) at 100 °C into 200 g water at 22 °C. Fill the m·c terms into heat lost = heat gained, then solve.

Tf = × 100 + × 22 + = °C
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Your turn — aluminum into water

Drop 80 g aluminum (c = 0.897) at 100 °C into 200 g water at 22 °C. Fill the m·c terms into heat lost = heat gained, then solve.

Tf = × 100 + × 22 + = °C
Tf = 71.76 × 100 + 836.8 × 2271.76 + 836.8 = 28.2 °C
Between 22 and 100, and close to 22: the water's m·c of 836.8 far outweighs the aluminum's 71.76. ✓
Dr. Karmach

Worked example 3 — enthalpy from a temperature rise

q = m·c·ΔT, then ΔH = −q / moles
given: 100.0 g solution · 20.0 → 26.8 °C · 0.0500 mol reacted · wanted: ΔH per mole

An acid and a base neutralize in a cup holding 100.0 g of solution. The temperature climbs from 20.0 to 26.8 °C, and 0.0500 mol of water forms. Find ΔH per mole.

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Worked example 3 — solution

q = m·c·ΔT, then ΔH = −q / moles
100.0 g solution · ΔT = +6.8 °C · 0.0500 mol

Two moves are needed.

Step 1 · Heat gained by the solution

q = 100.0 g × 4.184 J/g·°C × 6.8 °C = 2845 J
Dr. Karmach

Worked example 3 — solution

q = m·c·ΔT, then ΔH = −q / moles
100.0 g solution · ΔT = +6.8 °C · 0.0500 mol
Two moves are needed. Step 1 · Heat gained by the solution
q = 100.0 g × 4.184 J/g·°C × 6.8 °C = 2845 J
Step 2 · Heat per mole of reaction
ΔH = −2845 J0.0500 mol = −56,900 J/mol = −56.9 kJ/mol
The solution warmed, so the reaction released heat: ΔH is negative, exothermic. ✓
Dr. Karmach

Where this goes wrong

100 g iron at 90 °C into 150 g water at 20 °C → Tf = 24.7 °C
the correct final temperature both bodies reach
Ignoring the water's share. Assuming the metal just cools to the water's temperature gives Tf = 20 °C. The water warms too; its gain is part of the balance.
Adding the two heats. Same sign on both sides gives Tf = 14.6 °C, below both starting temperatures. One body loses heat, the other gains it: heat lost = heat gained.
Swapping the masses onto the wrong specific heats. Pairing 150 with the iron and 100 with the water gives Tf = 29.7 °C. Each mass keeps its own substance's c.
Dr. Karmach

Practice 1

heat lost = heat gained
given: 60 g copper at 120 °C · 125 g water at 25 °C · c: copper 0.385, water 4.184 · wanted: Tf

A 60 g copper block at 120 °C drops into 125 g of water at 25 °C. What final temperature do they reach?

  1. 40.3 °C
  2. 29.0 °C
  3. 25.0 °C
  4. 20.6 °C
Dr. Karmach

Practice 1 — answer: B

heat lost = heat gained
60 g copper at 120 °C · 125 g water at 25 °C
Tf = 23.1 × 120 + 523 × 2523.1 + 523 = 29.0 °C — answer B

A swapped the masses onto the wrong specific heats: 40.3 °C. C ignored the water's share and let the copper cool to 25 °C. D added the heats with the same sign: 20.6 °C, below the water's own start.

29.0 °C sits between 25 and 120, close to the water: its m·c of 523 outweighs the copper's 23.1. ✓
Dr. Karmach

Practice 2

heat lost = heat gained
given: 100 g silver at 90 °C · 150 g water at 20 °C · c: silver 0.235, water 4.184 · wanted: Tf

A 100 g silver coin at 90 °C drops into 150 g of water at 20 °C. What final temperature do they reach?

  1. 25.4 °C
  2. 22.5 °C
  3. 20.0 °C
  4. 17.3 °C
Dr. Karmach

Practice 2 — answer: B

heat lost = heat gained
100 g silver at 90 °C · 150 g water at 20 °C
Tf = 23.5 × 90 + 627.6 × 2023.5 + 627.6 = 22.5 °C — answer B

A swapped the masses onto the wrong specific heats: 25.4 °C. C ignored the water's share and let the silver cool to 20 °C. D added the heats with the same sign: 17.3 °C, below both starting temperatures.

22.5 °C sits between 20 and 90, close to the water: silver's small m·c of 23.5 barely moves it. ✓
Dr. Karmach

Check yourself

  1. A hot metal is dropped into cool water in an insulated cup. Write the conservation law that relates the heat the metal loses to the heat the water gains.
  2. A final temperature comes out below both starting temperatures. What went wrong, and what does the correct answer always sit between?

A reaction releases a fixed amount of heat per mole, measured here as ΔH. That heat travels with the balanced equation, and reversing or scaling the equation changes ΔH in step: thermochemical equations.

Dr. Karmach

6 · Thermochemical Equations

Treat ΔH as tied to the equation as written — reverse the equation and flip its sign, scale the coefficients and scale ΔH in step, and use ΔH as a conversion factor to find the heat released or absorbed by a given amount of substance.

Dr. Karmach

The gas bill charges for heat

The utility charges for energy delivered, not the gas itself. Burning a set amount of fuel releases a set amount of heat, and the bill counts that heat.

Dr. Karmach

ΔH is tied to the equation as written

A thermochemical equation pairs a balanced equation with its ΔH. That ΔH belongs to those exact coefficients and states. Change the equation, and ΔH changes with it.

CH₄(g) + 2 O₂(g) → CO₂(g) + 2 H₂O(l) · ΔH = −890 kJ
−890 kJ is released when 1 mol CH₄ burns exactly as written
Dr. Karmach

Reverse the equation, flip the sign

Running a reaction backward reverses its heat flow. A release becomes an equal absorption. The magnitude is unchanged; only the sign flips.

H₂O(l) → H₂(g) + ½ O₂(g) · ΔH = +286 kJ
forward, ΔH = −286 kJ (released) · reversed, ΔH = +286 kJ (absorbed)
Dr. Karmach

Scale the coefficients, scale ΔH

ΔH is proportional to the amount reacting. Double every coefficient and twice as much reacts, so ΔH doubles. Halve them and ΔH halves.

2 H₂(g) + O₂(g) → 2 H₂O(l) · ΔH = −572 kJ
from H₂ + ½ O₂ → H₂O, ΔH = −286 kJ, scaled ×2
Dr. Karmach

ΔH is a conversion factor

The coefficients turn ΔH into a factor: kJ per mole of any species in the equation. Chain it with molar mass to move between grams of fuel and kilojoules of heat.

Dr. Karmach

The method

  1. Match the target to the given. Reverse the equation if the target runs backward.
  2. Flip the sign for a reversal; scale ΔH by the factor that scales the coefficients.
  3. For heat, use ΔH as kJ per mole and cancel units.
Dr. Karmach

Worked example 1 — reversing an equation

C(s) + O₂(g) → CO₂(g) · ΔH = −394 kJ
given: this equation and ΔH · wanted: ΔH for the reverse

Carbon burns to carbon dioxide, releasing 394 kJ. What is ΔH for the reverse, CO₂(g) → C(s) + O₂(g)?

Dr. Karmach

Worked example 1 — solution

C(s) + O₂(g) → CO₂(g) · ΔH = −394 kJ
forward: 394 kJ released

Step 1 · Match the target to the given

The target runs the other way: CO₂ breaks apart into C and O₂. It is the reverse.

Dr. Karmach

Worked example 1 — solution

C(s) + O₂(g) → CO₂(g) · ΔH = −394 kJ
forward: 394 kJ released
Step 1 · Match the target to the given Step 2 · Flip the sign
CO₂(g) → C(s) + O₂(g) · ΔH = +394 kJ
−394 kJ released → +394 kJ absorbed · same magnitude, opposite sign
Splitting CO₂ must cost exactly the energy its formation released. ✓
Dr. Karmach

Worked example 2 — reverse and scale

H₂(g) + ½ O₂(g) → H₂O(l) · ΔH = −286 kJ
given: this equation and ΔH · wanted: ΔH for 2 H₂O(l) → 2 H₂(g) + O₂(g)

Find ΔH for 2 H₂O(l) → 2 H₂(g) + O₂(g).

A common first attempt: flip the sign but leave the coefficients' change out of ΔH. Test it.

Dr. Karmach

Worked example 2 — solution

H₂(g) + ½ O₂(g) → H₂O(l) · ΔH = −286 kJ
target: 2 H₂O(l) → 2 H₂(g) + O₂(g)

Step 1 · Match the target to the given

The target is reversed and every coefficient is doubled. Two changes, so ΔH takes two steps.

Dr. Karmach

Worked example 2 — solution

H₂(g) + ½ O₂(g) → H₂O(l) · ΔH = −286 kJ
target: 2 H₂O(l) → 2 H₂(g) + O₂(g)
Step 1 · Match the target to the given Step 2 · Flip the sign, then scale ΔH
ΔH = −(2 × −286 kJ) = +572 kJ
Flip only: +286 kJ ✗. Scale only: −572 kJ ✗. Both moves give +572 kJ: forming 2 mol water released 572 kJ, so splitting it absorbs 572 kJ. ✓
Dr. Karmach

Take-home: reverse flips, scale multiplies

reverse the equation → flip the sign of ΔH
scale the coefficients by n → multiply ΔH by n · do both when both apply

Each change to the equation changes ΔH. Reversing flips the sign. Scaling multiplies. A target that is both reversed and doubled needs both moves.

Dr. Karmach

Your turn — reverse and scale

Given 2 SO₂(g) + O₂(g) → 2 SO₃(g), ΔH = −198 kJ, find ΔH for 4 SO₃(g) → 4 SO₂(g) + 2 O₂(g) (reversed and doubled).

ΔH = ( × −198 kJ) = kJ
Dr. Karmach

Your turn — reverse and scale

Given 2 SO₂(g) + O₂(g) → 2 SO₃(g), ΔH = −198 kJ, find ΔH for 4 SO₃(g) → 4 SO₂(g) + 2 O₂(g) (reversed and doubled).

ΔH = ( × −198 kJ) = kJ
ΔH = −(2 × −198 kJ) = +396 kJ
Reversed, so the sign flips; doubled, so ×2. Forming 4 mol SO₃ released 396 kJ, so the reverse absorbs it. ✓
Dr. Karmach

Worked example 3 — grams of fuel to kilojoules

CH₄(g) + 2 O₂(g) → CO₂(g) + 2 H₂O(l) · ΔH = −890 kJ
given: 40.0 g CH₄ burned · molar mass CH₄ = 16.04 g/mol · wanted: ΔH

Burn 40.0 g of methane completely. What is ΔH for this amount of fuel?

Two conversion factors are needed: grams to moles, then moles to kilojoules.

Dr. Karmach

Worked example 3 — solution

CH₄(g) + 2 O₂(g) → CO₂(g) + 2 H₂O(l) · ΔH = −890 kJ
40.0 g CH₄ · 16.04 g/mol · −890 kJ per 1 mol CH₄

Two conversion factors are needed.

Step 1 · Pick the ΔH orientation

Two orientations exist. Only the one canceling mol CH₄ is used: −890 kJ / 1 mol CH₄, not 1 mol CH₄ / −890 kJ.

Dr. Karmach

Worked example 3 — solution

CH₄(g) + 2 O₂(g) → CO₂(g) + 2 H₂O(l) · ΔH = −890 kJ
40.0 g CH₄ · 16.04 g/mol · −890 kJ per 1 mol CH₄
Two conversion factors are needed. Step 1 · Pick the ΔH orientation Step 2 · Chain grams → moles → kilojoules
40.0 g × 1 mol16.04 g × −890 kJ1 mol = −2220 kJ
40.0 g is about 2.5 mol, each releasing 890 kJ: ΔH ≈ −2220 kJ, about 2200 kJ given off. ✓
Dr. Karmach

Where this goes wrong

from H₂ + ½ O₂ → H₂O(l), ΔH = −286 kJ, find 2 H₂O(l) → 2 H₂ + O₂
correct: reverse and double → ΔH = +572 kJ
Forgetting to flip the sign. Scaling but keeping the original sign gives −572 kJ. The target is the reverse, so the sign must flip: +572 kJ.
Forgetting to scale. Flipping the sign but leaving the coefficients out gives +286 kJ. Every coefficient doubled, so ΔH doubles: +572 kJ.
Leaving ΔH unchanged. Copying −286 kJ ignores both moves. The equation was reversed and doubled; ΔH must be too.
Dr. Karmach

Practice 1

N₂(g) + O₂(g) → 2 NO(g) · ΔH = +180 kJ
given: this equation and ΔH · wanted: ΔH for 4 NO(g) → 2 N₂(g) + 2 O₂(g)

What is ΔH for 4 NO(g) → 2 N₂(g) + 2 O₂(g) (the reverse, doubled)?

  1. +360 kJ
  2. −360 kJ
  3. −180 kJ
  4. +180 kJ
Dr. Karmach

Practice 1 — answer: B

N₂(g) + O₂(g) → 2 NO(g) · ΔH = +180 kJ
target: 4 NO → 2 N₂ + 2 O₂ · reversed and ×2
ΔH = −(2 × 180 kJ) = −360 kJ — answer B

A scaled but kept the sign: 2 × 180 = +360 kJ. C flipped the sign but did not scale: −180 kJ. D left ΔH untouched at +180 kJ.

Forming 2 mol NO absorbed 180 kJ, so making 4 mol the reverse way releases twice that: −360 kJ. ✓
Dr. Karmach

Practice 2

C₃H₈(g) + 5 O₂(g) → 3 CO₂(g) + 4 H₂O(l) · ΔH = −2220 kJ
given: ΔH per mole of propane · wanted: ΔH for 0.500 mol

What is ΔH when 0.500 mol of propane burns?

  1. −1110 kJ
  2. +1110 kJ
  3. −4440 kJ
  4. −2220 kJ
Dr. Karmach

Practice 2 — answer: A

C₃H₈(g) + 5 O₂(g) → 3 CO₂(g) + 4 H₂O(l) · ΔH = −2220 kJ
0.500 mol propane · −2220 kJ per 1 mol
0.500 mol × −2220 kJ1 mol = −1110 kJ — answer A

B dropped the negative sign: combustion releases heat, so ΔH is negative. C divided by 0.500 instead of multiplying: −4440 kJ. D reported the per-mole value and ignored the 0.500 mol.

Half a mole releases half of 2220 kJ, and the sign stays negative: −1110 kJ. ✓
Dr. Karmach

Check yourself

  1. An equation is reversed and its coefficients tripled. What two changes apply to ΔH?
  2. A reaction releases 500 kJ per mole. Write the conversion factor that turns moles of it into kilojoules, and state which unit cancels.

A single reaction's ΔH can be reversed and scaled. Adding several such equations, each reversed or scaled to line up, builds the ΔH of a reaction never measured directly: Hess's law.

Dr. Karmach

Can you…?

  • ☐ distinguish heat from work and convert among joules, calories, and Calories?
  • ☐ identify exothermic and endothermic processes from equations, energy diagrams, and the sign of ΔH?
  • ☐ apply q = m·c·ΔT to find heat, mass, specific heat, or temperature change?
  • ☐ compare how specific heats set different substances' temperature response to the same heat?
  • ☐ use calorimetry data to determine the heat of a process or reaction?
  • ☐ reverse and scale thermochemical equations and use ΔH as a conversion factor?

If any box stays empty, the practice site has a drill for it. 🧪

Dr. Karmach