How do you add warming and melting in one problem?
You will be able to: Partition a thermal path into stages and add their signed heat contributions.
How do you add warming and melting in one problem?
Turning cold ice into warm water requires warming the ice, melting it, and warming the liquid. Skipping the middle job can miss most of the energy.
A useful starting point: Why does a heating curve have slopes and plateaus? →
Words and symbols before equations
- Thermal path
- An ordered sequence connecting stated initial and final conditions.
- Stage
- A part using one appropriate heat relation.
- Phase-specific heat capacity
- A c value that applies to the named phase.
- Total heat
- Sum of all signed contributions in the stated path.
What this picture assumes
Ice begins at −10 °C and ends fully melted at selected liquid temperature, approximately 1 atm. Supplied M=18.0 g/mol, c_ice=2.00, c_water=4.00 J/(g·K), ΔHfus=6.00 kJ/mol. No vaporization or heat loss.
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.
- Total q=7.8 kJ. Mass scales all three terms; final liquid temperature changes only the last term. No vaporization is needed for this target.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the chemistry
Identify initial and final phases and temperatures before choosing equations. Mark each transition temperature the path crosses.
Within a phase use mcΔT with that phase’s c; at a transition use nΔH with its signed direction. Convert mass to moles when necessary.
Add in one common energy unit. Heating and cooling paths reverse signs when the states are reversed under the same conditions.
Check the resulting path: final water at 20 °C does not need a vaporization step. Conversely, obtaining steam from liquid does require that step.
A worked example, step by step
Convert 18.0 g ice at −10 °C into water at 20 °C. Use M=18.0 g/mol, c_ice=2.00 J/(g·K), ΔHfus=6.00 kJ/mol and c_water=4.00 J/(g·K), all supplied rounded values.
- Warm ice: 18.0×2.00×10=360 J=0.360 kJ.
- Melt: n=18.0/18.0=1.00 mol; q=1.00×6.00=6.00 kJ.
- Warm liquid: 18.0×4.00×20=1440 J=1.44 kJ.
- Total=0.360+6.00+1.44=7.80 kJ absorbed. The final state is liquid, so no vaporization is included.
One mcΔT across different phases hides phase-change energy and uses the wrong heat capacity for some segments.
If final liquid temperature rises but mass stays fixed, does melting heat change?
Compare with an explanation
No. The same amount still melts; only the liquid-warming term changes.
Predict. Change one thing. Explain.
Change the mass and final liquid temperature while the ice starts at −10 °C. Compare the warming and melting contributions; predict which term changes with final temperature.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
Total q=7.8 kJ. Mass scales all three terms; final liquid temperature changes only the last term. No vaporization is needed for this target.
Ice begins at −10 °C and ends fully melted at selected liquid temperature, approximately 1 atm. Supplied M=18.0 g/mol, c_ice=2.00, c_water=4.00 J/(g·K), ΔHfus=6.00 kJ/mol. No vaporization or heat loss.
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 questionUsing the same supplied constants, convert 9.00 g ice at −10 °C to water at 20 °C. Show each heat term and the total.
This response is not submitted or saved. Copy it before leaving.
Compare with the answer and four-point rubric
- 1 point: Ice warming=9.00×2.00×10=180 J=0.180 kJ.
- 1 point: Fusion=(9.00/18.0)×6.00=3.00 kJ.
- 1 point: Liquid warming=9.00×4.00×20=720 J=0.720 kJ.
- 1 point: Total=3.90 kJ absorbed, half the 18 g result.
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 1What comes before calculation?
Identify every phase and temperature segment.
RECALL 2Why convert J and kJ?
Terms can only be added in consistent units.
RECALL 3Which c should be used?
The value for the phase in that stage.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
How do you add warming and melting in one problem?
- q_total=Σq_stage, with common units.
- Each stage uses either mcΔT or nΔH_transition under its stated conditions.
Remember: One mcΔT across different phases hides phase-change energy and uses the wrong heat capacity for some segments.
Conditions: Ice begins at −10 °C and ends fully melted at selected liquid temperature, approximately 1 atm. Supplied M=18.0 g/mol, c_ice=2.00, c_water=4.00 J/(g·K), ΔHfus=6.00 kJ/mol. No vaporization or heat loss.
Refresh Kid · AP Chemistry Unit 6 · Objectives 6.5.A · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 6.5, objective 6.5.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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