Find the shared final temperature
You will be able to: Use energy conservation to analyze two bodies reaching thermal equilibrium.
Why is the final temperature not always halfway between the starts?
Place a small hot metal sample in cooler water. Both approach a common temperature, but it need not be the arithmetic average. The water and metal may need very different amounts of energy for the same temperature change.
A useful starting point: How much energy changes a temperature? →
Words and symbols before equations
- Calorimetry
- Using temperature changes and energy conservation to infer energy transfer or material properties.
- Insulated combined system
- An approximation in which negligible energy escapes the bodies plus their container.
- Signed heat Q_i
- mc(T_f−T_i) for each body; positive if it warms.
- Common T_f
- Final equilibrium temperature, shared by bodies in thermal contact.
What this picture assumes
Hot body C=100 J/K at 80 °C; cold body starts at 20 °C. Insulated together, negligible cup heat capacity/work, constant capacities, no phase changes. Final temperature is a weighted energy balance.
Connect the picture to the physics
Choose both bodies together as the system. If no energy leaves, no significant work occurs and the container’s heat capacity is negligible, their thermal transfers add to zero: m₁c₁(T_f−T₁)+m₂c₂(T_f−T₂)=0.
Solving gives T_f=(C₁T₁+C₂T₂)/(C₁+C₂), where C_i=m_i c_i. This is a weighted average. With positive heat capacities and no other source, the final temperature lies between the initial values and nearer the initial temperature of the body with larger heat capacity.
A real calorimeter and thermometer absorb energy too. Include their heat capacities when significant, and discuss energy leakage. A measured result outside the predicted interval signals a missing transfer, a phase change, a measurement error or an invalid model.
A worked example, step by step
Body A has C_A=100 J/K at 80 °C. Body B has C_B=300 J/K at 20 °C. Find T_f and energy transferred in an ideal insulated setup.
- Set 100(T_f−80)+300(T_f−20)=0.
- 400T_f=14000, so T_f=35 °C.
- A: Q_A=100(35−80)=−4500 J. B: Q_B=300(35−20)=+4500 J.
- The result lies closer to 20 °C because B has the larger heat capacity.
An unweighted average works only when heat capacities are equal. Include the cup if its energy change is significant.
Can the ideal two-body result be hotter than both initial temperatures?
Compare with an explanation
No, not with positive heat capacities, no phase changes and no additional energy source.
Predict. Change one thing. Explain.
Hold the initial temperatures at 80 °C and 20 °C. Change the cold body’s heat capacity. Predict which initial temperature the result approaches and check that energy gained equals energy lost.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
Final T=35 °C. Q_hot=-4500 J and Q_cold=4500 J; their sum is zero. Cold-body heat capacity=300 J/K; hot-body heat capacity=100 J/K.
Hot body C=100 J/K at 80 °C; cold body starts at 20 °C. Insulated together, negligible cup heat capacity/work, constant capacities, no phase changes. Final temperature is a weighted energy balance.
Explain what you noticed: Which quantity changed? Which stayed fixed? Use the relevant particle, temperature or energy relationship to justify your prediction.
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 questionTwo insulated bodies have C_A=200 J/K at 60 °C and C_B=100 J/K at 30 °C. (a) Write the signed balance. (b) Find T_f. (c) Find each heat transfer. (d) Explain how a cool cup with appreciable heat capacity affects the result.
This response is not submitted or saved. Copy it before leaving.
Compare with the answer and four-point rubric
- 1 point: 200(T_f−60)+100(T_f−30)=0.
- 1 point: T_f=50 °C.
- 1 point: Q_A=−2000 J and Q_B=+2000 J.
- 1 point: A cup initially at 30 °C absorbs some energy and lowers T_f relative to the negligible-cup model.
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 use signed Q for each body?
It makes heating and cooling cancel in the combined energy balance.
RECALL 2Which temperature has more weight?
The initial temperature of the body with larger mc.
RECALL 3What real equipment may need inclusion?
The calorimeter, thermometer and any significant energy leakage.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Find the shared final temperature
- ΣQ=0 for the stated insulated no-work model.
- Q_i=m_i c_i(T_f−T_i).
- T_f=(C₁T₁+C₂T₂)/(C₁+C₂).
Remember: An unweighted average works only when heat capacities are equal. Include the cup if its energy change is significant.
Conditions: Hot body C=100 J/K at 80 °C; cold body starts at 20 °C. Insulated together, negligible cup heat capacity/work, constant capacities, no phase changes. Final temperature is a weighted energy balance.
Refresh Kid · AP Physics 2 Unit 1 (official Unit 9) · Objectives 9.5.A · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 9.5, objectives 9.5.A. CED effective Fall 2024, current PDF ©2026; checked September 16, 2026. Refresh Kid calls this the first AP Physics 2 unit; College Board numbers it Unit 9, continuing after AP Physics 1. The lesson breakdown and questions are original Refresh Kid work, not official topic subdivisions.
Implementation and automated checks are separate from independent teacher review and observation of students. Both human review stages remain pending. This is a review edition, not a certified or validated assessment.
Optional further resource: College Board’s released questions and scoring guides. Papers can combine units; this link is an archive, not an assignment of every question to this lesson.
Our learn, explore, practice and recall sequence is informed by the IES learning guide. The exact Refresh Kid implementation has not been evaluated for learning effectiveness.
Want to work through this with a tutor?
Bring your question about Find the shared final temperature. Your explanation and answers remain free to access.