Library / Heat transfer and thermal systems / HTS-46
Heat transfer and thermal systems
Heat stored through a complete melting interval
Energy to take a mass m from a solid at Ti below the melting point, through melting, to a liquid at Tf.
- Ref
- HTS-46
- Rev
- v1
- Field
- Heat transfer and thermal systems
- Result
- E [J]
The energy to take a mass m of material from solid at Ti, through melting at Tm, to liquid at Tf is E = m[cs(Tm − Ti) + Lf + cl(Tf − Tm)]. It sums the solid sensible heat, the latent heat and the liquid sensible heat, with a single melting temperature. Real materials often melt over a range and lose heat, so treat it as an upper bound.
Energy to take a mass m from a solid at Ti below the melting point, through melting, to a liquid at Tf. It adds solid sensible heat, latent heat and liquid sensible heat, with a single melting temperature and no losses.
Calculator
- Share of the energy that is latent heat76.92 %
- Share that is sensible heat23.08 %
- Energy per kilogram195,000 J/kg
What drives the result
Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.
- Melting and heating 2 kg through this interval stores 390,000 J (0.1083 kWh), 76.9 % of it as latent heat.
- The latent share is why phase-change stores beat plain water or rock in a narrow temperature window: the same energy would need a far larger temperature swing in a sensible store.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (J) | Local step | Change in result |
|---|---|---|---|---|---|---|
| m Mass | 0.01 to 1,000 kg | 1,950 | 1.95e8 | 1.95e8 | +10% | +10 % |
| Lf Latent heat of fusion | 10,000 to 500,000 J/kg | 110,000 | 1,090,000 | 980,000 | +10% | +7.69 % |
| Tf Final (liquid) temperature | 310 to 400 K | 390,000 | 840,000 | 450,000 | +10% | +39.7 % |
| Ti Initial (solid) temperature | 200 to 300 K | 350,000 | 750,000 | 400,000 | +10% | — |
| cs Solid specific heat | 500 to 5,000 J/(kg K) | 360,000 | 450,000 | 90,000 | +10% | +1.03 % |
| cl Liquid specific heat | 500 to 5,000 J/(kg K) | 350,000 | 440,000 | 90,000 | +10% | +1.28 % |
| Tm Melting temperature | 300 to 310 K | 380,000 | 390,000 | 10,000 | +10% | — |
Sweep of Latent heat of fusion
| Latent heat of fusion (J/kg) | Stored energy (J) |
|---|---|
| 10,000 | 110,000 |
| 71,250 | 232,500 |
| 132,500 | 355,000 |
| 193,800 | 477,500 |
| 255,000 | 600,000 |
| 316,300 | 722,500 |
| 377,500 | 845,000 |
| 438,800 | 967,500 |
| 500,000 | 1,090,000 |
Sweep one variable
Variables and how each one contributes
Stored energy
What the equation gives you. Everything else feeds it.
Mass
The mass of the storage material, in kg. Raises the stored energy in direct proportion. Raising it by 10 % (to 2.2) moves the result by +10 %, from 390,000 to 429,000 J.
Solid specific heat
The solid specific heat, in J/(kg K). Raises the solid sensible heat. Raising it by 10 % (to 2,200) moves the result by +1.03 %, from 390,000 to 394,000 J.
Melting temperature
The melting temperature, in K. A higher melting point increases the solid heating and reduces the liquid heating. Raising it by 10 % (to 330) gives no valid result at the other default values.
Initial (solid) temperature
The initial solid temperature, in K. A colder start raises the solid sensible heat. Raising it by 10 % (to 319) gives no valid result at the other default values.
Latent heat of fusion
The latent heat of fusion, in J/kg. Raises the stored energy in direct proportion to the latent term. Raising it by 10 % (to 165,000) moves the result by +7.69 %, from 390,000 to 420,000 J.
Liquid specific heat
The liquid specific heat, in J/(kg K). Raises the liquid sensible heat. Raising it by 10 % (to 2,750) moves the result by +1.28 %, from 390,000 to 395,000 J.
Final (liquid) temperature
The final liquid temperature, in K. A hotter final state raises the liquid sensible heat. Raising it by 10 % (to 341) moves the result by +39.7 %, from 390,000 to 545,000 J.
Worked example with the values above
- Solid heating: cs(Tm − Ti) = 2,000 × 10 = 20,000 J/kg; latent: 150,000 J/kg.
- Liquid heating: cl(Tf − Tm) = 2,500 × 10 = 25,000 J/kg.
- E = 2 × 195,000 = 390,000 J.
Assumptions and limits
- Constant properties in each phase, a single melting temperature and no losses unless stated.
- Steady or quasi-steady balances, with absolute temperatures where logarithms or powers are used.
Common mistakes
- Forgetting the latent heat term, or counting it twice.
- Using °C in an entropy or radiation formula.
- Using these simple formulas when the Stefan number is not small.
Statistics and uncertainty
Latent heats and specific heats are tabulated to a few per cent, but real storage materials melt over a range and lose heat, so real systems typically deliver 5 % to 20 % less than the ideal values.
Where it is used
- Sizing thermal storage, ice banks and phase-change units.
- Checking exchanger data, mixing streams and cooling times.
Maths you need
- Latent heat: energy per kilogram to change phase
- Heat-capacity rate: C = ṁcp
- Second law: entropy generation cannot be negative