Why is the final temperature not always the average?
You will be able to: Explain how unequal heat capacities determine the common equilibrium temperature.
Why is the final temperature not always the average?
Mixing a little hot water with a lot of cool water leaves a temperature closer to the cool water’s start. The same energy changes the smaller portion’s temperature more.
A useful starting point: Why does heat flow from warm to cool? →
Words and symbols before equations
- Heat capacity, C
- Energy required to change a whole sample’s temperature by one degree; J/K.
- Equilibrium temperature, Tf
- Common final temperature after net heat exchange stops.
- Isolated pair
- Two bodies exchanging heat with each other but negligibly with anything else.
What this picture assumes
Insulated pair, hot C=100 J/K at 70 °C, cool body initially 30 °C. Constant heat capacities, no phase change or apparatus heat. Final state only; no elapsed-time prediction.
Read the picture in three steps
- Read the species and labels first. Identify what each symbol and line represents. Read the units and fixed conditions before comparing quantities.
- Tf=40 °C. Hot body loses 3000 J and cool body gains the same amount. A greater cool-body capacity anchors Tf closer to 30 °C.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the chemistry
For constant heat capacities and no phase changes, each body has q=C(Tf−Ti). The initially hot body has negative q; the cool body positive q.
Conservation requires C_hot(Tf−T_hot)+C_cool(Tf−T_cool)=0. Solving gives a heat-capacity-weighted average.
The simple arithmetic average works only when the two heat capacities are equal. A larger capacity anchors the final temperature closer to its own initial value.
This equilibrium calculation predicts the endpoint, not how long cooling takes. A cup, thermometer or room exchanging heat would require additional terms.
A worked example, step by step
Two insulated bodies have C_hot=100 J/K at 70 °C and C_cool=300 J/K at 30 °C. Find Tf and check the transfers.
- Set 100(Tf−70)+300(Tf−30)=0.
- Collect terms: 400Tf=16000, so Tf=40 °C.
- Hot body: q=100(40−70)=−3000 J; cool body: q=300(40−30)=+3000 J.
- The signed sum is zero and Tf is closer to 30 °C because that body has greater heat capacity.
Do not average temperatures without considering heat capacities and possible phase changes.
What happens when the two capacities are equal?
Compare with an explanation
The final temperature is their arithmetic average under this model’s assumptions.
Predict. Change one thing. Explain.
Keep starting temperatures at 70 and 30 °C. Increase the cool body’s heat capacity and predict where the final temperature moves; check equal and opposite heat.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
Tf=40 °C. Hot body loses 3000 J and cool body gains the same amount. A greater cool-body capacity anchors Tf closer to 30 °C.
Insulated pair, hot C=100 J/K at 70 °C, cool body initially 30 °C. Constant heat capacities, no phase change or apparatus heat. Final state only; no elapsed-time prediction.
Explain what you noticed: Answer the investigation prompt above. State one observation and explain it using heat-flow signs, energy conservation, phase changes, bond inventories or the stated thermochemical path. Identify what the representation cannot tell you.
Apply the idea to a fresh problem Practice →Show what you understand.
Two original questions are a starting check, not proof of mastery. Explain your choice before revealing the answer.
Original written challenge
4 points · self-check · not an official AP questionC1=200 J/K at 80 °C and C2=200 J/K at 20 °C exchange heat in an isolated pair. Find Tf, each q and one condition that would invalidate the simple result.
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Compare with the answer and four-point rubric
- 1 point: Equal capacities give Tf=50 °C.
- 1 point: q1=200(50−80)=−6000 J.
- 1 point: q2=200(50−20)=+6000 J.
- 1 point: A phase change, heat leak or significant container heat capacity would require a revised balance.
Accept equivalent correct methods and explanations. This is a Refresh Kid teaching rubric, not an official AP scoring guideline.
Retrieve it before you reveal it.
RECALL 1Why a weighted average?
Each degree of temperature change involves a different amount of energy if capacities differ.
RECALL 2What must the signed heat sum be?
Zero for the isolated pair.
RECALL 3Does this model predict elapsed time?
No; it predicts the equilibrium endpoint.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Why is the final temperature not always the average?
- Tf=(C1T1+C2T2)/(C1+C2), for an insulated pair with constant C and no phase change.
- q1+q2=0.
Remember: Do not average temperatures without considering heat capacities and possible phase changes.
Conditions: Insulated pair, hot C=100 J/K at 70 °C, cool body initially 30 °C. Constant heat capacities, no phase change or apparatus heat. Final state only; no elapsed-time prediction.
Refresh Kid · AP Chemistry Unit 6 · Objectives 6.3.A · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 6.3, objective 6.3.A. CED effective Fall 2024 and June 2026 clarifications checked September 16, 2026. Unit 6: Thermochemistry, Topics 6.1–6.9. Focused lesson names, examples, models and assessments are original Refresh Kid teaching materials, not additional official topics or official AP questions. Official corrections.
The model states its assumptions beside the diagram. Technical enthalpy/internal-energy distinctions and formal state-function terminology are not assessed in the current AP framework. Constant-pressure heat, conservation, phase-specific capacities, reaction amounts and Hess sums are taught here with explicit conditions. Supplied rounded data and original molecular geometry are teaching models, not experimental measurements. A phase transition preserves molecular identity; a bond-energy accounting path is not an actual reaction mechanism.
Teaching resources: The Organic Chemistry Tutor video titles/descriptions and topic coverage were checked for optional links; no claim is made to have watched every video. No creator scripts, examples, worksheets or artwork were copied. GitHub’s 3D website collection and its Three.js camera-control example informed the idea of controllable spatial inspection. Scientific diagrams, geometry and interactions here are original. The self-hosted Three.js runtime retains its MIT license. Camera rotation changes the view, not the chemistry.
Independent teacher review and observation of students remain pending. Implementation checks do not certify scientific accuracy, accessibility or learning effectiveness. This is a review edition.
Optional official resource: Released AP Chemistry questions and scoring guides. This archive contains questions across units; it is not an assignment of every question to this lesson.
The teaching sequence is informed by the IES learning guide; this exact implementation has not been evaluated with learners.
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