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LESSON 15 / 16 · TOPIC 9.5

Measure specific heat with a graph

You will be able to: Design and interpret a heating experiment, including systematic energy losses.

Official College Board Unit 9Free study resourceReview editionTeacher review pending

How can a slope reveal a material property?

A 0.5 kg block warms by 2 K after receiving 1000 J and by 4 K after receiving 2000 J. Plot several energy inputs against their temperature rises. The slope connects the visible trend to the sample’s heat capacity.

A useful starting point: What sets the rate through a wall? →

Words and symbols before equations

Independent variable
Quantity deliberately changed, here cumulative energy supplied.
Best-fit slope
Trend through several data points, not automatically a line joining two noisy points.
Heater power P_h
Measured energy input per second; E_in=P_h t when constant.
Systematic bias
A repeated shift caused by a model or measurement issue, such as energy loss to surroundings.
Synthetic heating data · input versus observed riseHeater input energy (J)Temperature rise (K)002.51100522007.53300104400
Read this model snapshot. Slope=400 J/K. Ignoring 0% input loss gives c_apparent=800 J/(kg·K); true c=800 J/(kg·K). All data points are calculated, not measured.
What this picture assumes

Synthetic heating data: sample mass 0.5 kg and true c=800 J/(kg·K). A constant chosen fraction of input fails to warm the sample; no random noise or phase change. Real losses may vary and produce curvature. Points are predictions, not measurements.

Connect the picture to the physics

For ideal heating of only the sample, plot E_in on the vertical axis against ΔT on the horizontal. The slope is mc, so c=slope/m. If the axes are reversed, the slope becomes 1/(mc); record both axes and units before extracting c.

Measure sample mass, establish good thermal contact, insulate, distribute heat through the sample and allow the thermometer to represent its temperature. Record multiple energy/temperature pairs. Repeat trials and compare residuals; synthetic straight points do not establish the precision of real equipment.

If E_in includes energy absorbed by the heater, cup or surroundings, then E_in>mcΔT. Treating all input as sample heating generally overestimates c. In the simplified explorer, a fixed fraction of input is lost; real losses often change with temperature and time and can curve the graph.

A worked example, step by step

A 0.5 kg sample has an E_in-versus-ΔT best-fit slope of 500 J/K. Estimate c under negligible-loss assumptions. If 20% of input actually escaped, what is the true c in the constant-loss-fraction model?

  1. No-loss inference: c_app=slope/m=500/0.5=1000 J/(kg·K).
  2. Only 0.8E_in warmed the sample, so its actual heat capacity is 0.8(500)=400 J/K.
  3. c_true=400/0.5=800 J/(kg·K).
  4. Ignoring the lost energy overestimates c. This correction assumes a known constant fraction.
Common mix-up

Input energy is not automatically energy absorbed by the sample. A perfect synthetic line is not evidence of a perfect experiment.

CHECK THE IDEA

If less of the supplied energy warms the sample, how does the inferred c change when losses are ignored?

Compare with an explanation

It increases because a larger measured E_in is divided by the observed mΔT.

Now investigate one change Explore →

Predict. Change one thing. Explain.

Increase the fraction of input energy lost while keeping the true sample mass and specific heat fixed. Compare true and inferred c. All plotted points are generated by the chosen model, not measured data.

On narrow screens, swipe or scroll diagrams sideways to read all labels.

Synthetic heating data · input versus observed riseHeater input energy (J)Temperature rise (K)002.51100522007.53300104400

Slope=400 J/K. Ignoring 0% input loss gives c_apparent=800 J/(kg·K); true c=800 J/(kg·K). All data points are calculated, not measured.

Synthetic heating data: sample mass 0.5 kg and true c=800 J/(kg·K). A constant chosen fraction of input fails to warm the sample; no random noise or phase change. Real losses may vary and produce curvature. Points are predictions, not measurements.

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.

1. For a graph of E in J versus ΔT in K, slope has units…

Show answer and reasoning

J/K. Slope is energy divided by temperature change; divide by mass to get c.

2. Ignoring heat absorbed by the cup tends to make inferred sample c…

Show answer and reasoning

Too high. Input includes energy that did not warm the sample, inflating E_in/(mΔT).

Original written challenge

4 points · self-check · not an official AP question

A 0.4 kg sample gives a best-fit E_in-versus-ΔT slope of 320 J/K. (a) Find apparent c. (b) Explain why several points are better than one. (c) Name a control or measurement needed. (d) Predict bias if 25% of E_in never warms the sample.

This response is not submitted or saved. Copy it before leaving.

Compare with the answer and four-point rubric
  1. 1 point: c_app=320/0.4=800 J/(kg·K).
  2. 1 point: Multiple points reveal scatter and possible nonlinearity and support a fitted trend.
  3. 1 point: Measure mass and actual heater energy, ensure uniform sample temperature, or control/estimate losses.
  4. 1 point: Ignoring that loss overestimates c; constant-fraction correction gives c_true=0.75(800)=600 J/(kg·K).

Accept equivalent correct methods and explanations. This is a Refresh Kid teaching rubric, not an official AP scoring guideline.

Recall the ideas without notes Review →

Retrieve it before you reveal it.

RECALL 1What is the slope of E versus ΔT?

Heat capacity mc under the ideal no-loss model.

RECALL 2How do you obtain specific heat?

Divide that slope by measured sample mass.

RECALL 3Why label synthetic data?

A calculated illustration cannot establish real experimental uncertainty or validate the model.

Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.

Measure specific heat with a graph

  • E_absorbed=mcΔT.
  • Plot E_in versus ΔT: c_app=slope/m if losses are neglected.
  • Constant loss fraction f: c_app=c_true/(1−f).

Remember: Input energy is not automatically energy absorbed by the sample. A perfect synthetic line is not evidence of a perfect experiment.

Conditions: Synthetic heating data: sample mass 0.5 kg and true c=800 J/(kg·K). A constant chosen fraction of input fails to warm the sample; no random noise or phase change. Real losses may vary and produce curvature. Points are predictions, not measurements.

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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