How much energy changes a temperature?
You will be able to: Use Q=mcΔT with a model that excludes phase changes and significant work.
Why do equal energy inputs warm materials by different amounts?
Give 1000 J to a 1 kg sample with specific heat 1000 J/(kg·K): it warms by 1 K. Give the same energy to a sample with twice that specific heat: it warms by only 0.5 K. The material property changes how much warming a transfer produces.
A useful starting point: Compare four thermodynamic processes →
Words and symbols before equations
- Specific heat c
- Energy per kilogram per kelvin temperature rise, in J/(kg·K).
- Heat capacity C=mc
- Energy per kelvin for the whole sample, in J/K.
- Temperature change ΔT
- Final minus initial; a 1 °C change equals a 1 K change.
- Phase change
- A change such as melting; Q=mcΔT alone does not describe latent energy transfer.
What this picture assumes
4000 J warms a sample initially at 20 °C. Constant c, negligible work, heat loss and apparatus heat capacity; no phase change. Curves show temperature rise versus energy absorbed. Settings represent different samples.
Connect the picture to the physics
For a sample of mass m with constant c, no phase change and negligible work, Q=mc(T_f−T_i). Heat entering gives positive ΔT; cooling gives negative Q. The magnitude depends on mass, material and temperature change.
Specific heat is a material property in this simplified model. Doubling mass doubles heat capacity, not specific heat. Different materials store microscopic energy differently because of their internal arrangement and interactions.
Use Celsius or kelvin differences consistently in this equation. Unlike a temperature ratio in the ideal gas law, the offset cancels in a difference. Treat c as independent of temperature here. For gases, the process and work must be specified before using a heat capacity.
A worked example, step by step
A 0.5 kg sample with c=800 J/(kg·K) receives 4000 J. It starts at 20 °C. Find its final temperature.
- C=mc=(0.5)(800)=400 J/K.
- ΔT=Q/(mc)=4000/400=10 K, also a 10 °C rise.
- T_f=20+10=30 °C.
- This assumes all the stated heat warms the sample, with no phase change or significant work.
Specific heat c and heat capacity mc are different. Q=mcΔT does not account for a phase change by itself.
Does a larger c mean a material warms more for a fixed Q and m?
Compare with an explanation
No. ΔT=Q/(mc), so a larger c gives a smaller rise.
Predict. Change one thing. Explain.
Change m and c while holding heat input at 4000 J. Predict which change gives the same temperature rise. Notice that larger heat capacity means a smaller rise for the same input.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
Heat capacity mc=800 J/K. Absorbing 4000 J raises T by 5 K from 20 °C to 25 °C.
4000 J warms a sample initially at 20 °C. Constant c, negligible work, heat loss and apparatus heat capacity; no phase change. Curves show temperature rise versus energy absorbed. Settings represent different samples.
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 questionA 2 kg sample cools from 40 °C to 30 °C and has c=500 J/(kg·K). (a) Find ΔT. (b) Find Q for the sample. (c) Find heat capacity. (d) State one assumption behind the calculation.
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Compare with the answer and four-point rubric
- 1 point: ΔT=−10 K.
- 1 point: Q=2(500)(−10)=−10000 J, energy leaves.
- 1 point: C=mc=1000 J/K.
- 1 point: Constant c, no phase change and negligible work; the stated transfer is for the sample.
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 1Units of c?
J/(kg·K).
RECALL 2Units of C?
J/K.
RECALL 3When is Q=mcΔT insufficient?
During phase change, or when other energy transfers and process-dependent behavior matter.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
How much energy changes a temperature?
- Q=mcΔT; C=mc.
- ΔT=Q/(mc).
- A temperature difference of 1 K equals a difference of 1 °C.
Remember: Specific heat c and heat capacity mc are different. Q=mcΔT does not account for a phase change by itself.
Conditions: 4000 J warms a sample initially at 20 °C. Constant c, negligible work, heat loss and apparatus heat capacity; no phase change. Curves show temperature rise versus energy absorbed. Settings represent different samples.
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.
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