SCIENCE · INVESTIGATE AND EXPLAIN

Test a phase boundary before solving a final temperature

Heat is energy transferred because of a temperature difference, not a substance stored inside a body. In a fixed phase with constant specific heat, the energy change is mcΔT. During an ideal phase change at specified pressure, mL can be transferred without changing temperature. A final-temperature equation must therefore be chosen only after checking which phases remain.

Learning outcome: Retain one partial-melting, one exact-boundary and one warmed-liquid case. Show the phase test before calculating temperature and close the energy ledger including the cup in every applicable term.

Starting recap

Declare temperature and time references, system boundaries and model regimes. Use absolute temperatures for ratios; distinguish state, path, oscillator phase and wave propagation.

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Try

With Ti=40 °C and ice mass 0.28 kg, available energy and melting requirement are both 84000 J: the final state is all liquid at 0 °C. Raising Ti to 50 °C leaves 21000 J to warm the liquid and cup.

Open the experiment

Predict

If available energy is insufficient to melt all ice in this packet, the equilibrium temperature is:

Your prediction and reason

Experiment

Retain one partial-melting, one exact-boundary and one warmed-liquid case. Show the phase test before calculating temperature and close the energy ledger including the cup in every applicable term.

Test melting capacity before predicting final temperature

Worked teaching model · synthetic values · Isolated hypothetical water/ice packet at fixed pressure: .5 kg liquid, ice initially 0 °C, cup heat capacity 100 J/K initially at liquid temperature; supplied c=4000 J/(kg K), latent heat 300000 J/kg. Neglect evaporation and temperature dependence. Constants are simplified teaching values, not measurements.

Initial warm-liquid temperature: 30 °C Initial ice mass: 0.14 kg

Reset keeps your prediction, captures and practice history.

Piecewise final temperature versus initial warm temperature

Unlike graph quantities rescale independently; read units and values, not screen angles. 1: phase-aware isolated balance (solid). Joined curves show the supplied model, not additional observations.

Equivalent numerical coordinates for Piecewise final temperature versus initial warm temperature

phase-aware isolated balance

initial warm temperature (°C)	final temperature (°C)
0	0
0.625	0
1.25	0
1.875	0
2.5	0
3.125	0
3.75	0
4.375	0
5	0
5.625	0
6.25	0
6.875	0
7.5	0
8.125	0
8.75	0
9.375	0
10	0
10.625	0
11.25	0
11.875	0
12.5	0
13.125	0
13.75	0
14.375	0
15	0
15.625	0
16.25	0
16.875	0
17.5	0
18.125	0
18.75	0
19.375	0
20	0
20.625	0.493421
21.25	0.986842
21.875	1.48026
22.5	1.97368
23.125	2.46711
23.75	2.96053
24.375	3.45395
25	3.94737
25.625	4.44079
26.25	4.93421
26.875	5.42763
27.5	5.92105
28.125	6.41447
28.75	6.90789
29.375	7.40132
30	7.89474
30.625	8.38816
31.25	8.88158
31.875	9.375
32.5	9.86842
33.125	10.3618
33.75	10.8553
34.375	11.3487
35	11.8421
35.625	12.3355
36.25	12.8289
36.875	13.3224
37.5	13.8158
38.125	14.3092
38.75	14.8026
39.375	15.2961
40	15.7895
40.625	16.2829
41.25	16.7763
41.875	17.2697
42.5	17.7632
43.125	18.2566
43.75	18.75
44.375	19.2434
45	19.7368
45.625	20.2303
46.25	20.7237
46.875	21.2171
47.5	21.7105
48.125	22.2039
48.75	22.6974
49.375	23.1908
50	23.6842

At most 82 representative computed coordinates including endpoints; the plotted curve can use a finer grid. A table of a revealed generator is not independent measurement evidence.

Calculated quantities under the declared measurement model
Quantity	Value	Unit
Energy released on cooling warm contents to zero	63000	J
Energy required to melt all initial ice	42000	J
All initial ice melted	1	flag
Melted ice mass	0.14	kg
Remaining ice mass	0	kg
Final equilibrium temperature	7.89474	°C
Final liquid mass	0.64	kg

Enough initial thermal energy melts all ice; any remainder warms the final liquid and cup.

The cup contributes 100 J/K to the initial warm side and to the final heat capacity. Test Qavailable against m_ice L before solving an all-liquid equation. At equality all initial ice just melts at 0 °C; above it, divide the leftover energy by the total liquid-plus-cup heat capacity. Supplied constants are simplified and no heat escapes in this packet.

Predict before inspecting the values. Save contrasting records with assumptions and one alternative explanation. This virtual display does not certify practical measurement technique.

Enough initial thermal energy melts all ice; any remainder warms the final liquid and cup.

Observation to attach to a capture

Captures preserve recipes and replay their calculated results. They are synthetic model evidence, not physical measurements.

Understand

Declare the state, system and observation

Build the algebraic model

The isolated original packet starts with 0.5 kg warm liquid and a cup of heat capacity 100 J/K at the same initial temperature. Ice starts at 0 °C. Supplied simplified values are c=4000 J/(kg K) and melting latent heat 300000 J/kg. Cooling the warm liquid and cup to zero releases (0.5×4000+100)Ti=2100Ti J. Compare this with miceL before assuming all ice melts.

Check the original evidence packet

At Ti=30 °C and ice mass 0.14 kg, available energy is 63000 J and complete melting requires 42000 J. The remaining 21000 J warms 0.64 kg liquid plus the cup, with total heat capacity 2660 J/K. Final temperature is 21000/2660≈7.89474 °C. At Ti=10 °C with the same ice mass, only 0.07 kg melts; 0.07 kg ice remains and equilibrium is at 0 °C.

Challenge the domain and transfer

At Ti=20 °C and ice mass 0.14 kg, all initial ice just melts at 0 °C. Equality is a phase boundary, not justification for a negative all-liquid temperature below it. Neglecting the cup would change both the initial available energy and final heat capacity. Heat leaking to the surroundings would require another signed transfer term. No actual hot liquid, ice handling or calorimeter construction is requested, and the constants are disclosed teaching approximations rather than measured values.

Worked original case

With Ti=40 °C and ice mass 0.28 kg, available energy and melting requirement are both 84000 J: the final state is all liquid at 0 °C. Raising Ti to 50 °C leaves 21000 J to warm the liquid and cup.

Bounded investigation

Retain one partial-melting, one exact-boundary and one warmed-liquid case. Show the phase test before calculating temperature and close the energy ledger including the cup in every applicable term.

Changed-case numerical practice

Try the changed case before feedback

Using the supplied latent heat 300000 J/kg, 18000 J is available to melt initially 0 °C ice. How much ice melts in kilograms if more ice is present than this energy can melt?

Numerical result in the requested units

This practice is public and unsaved. A matching number does not verify understanding. Retain your explanation, assumptions and changed-case evidence in the separate course workspace.

Sources and limits

Original content checked 30 October 2026 against OpenStax College Physics 2e. Packets are supplied or generated teaching evidence, not measured apparatus. No heating, pressure vessels, physical oscillators, sound exposure or construction is requested. Public checks and saved reports do not establish supervised practical skill, equipment safety or independent mastery.

Apply

Public practice checks a specific claim. It does not establish independent mastery.

Which error results from applying a liquid-only temperature equation before checking melting capacity?

Your explanation and limitations

Continue

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Previous: Connect an ideal-gas state to a molecular energy distribution Next: Infer an interface temperature from a steady heat ledger

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