Example Data Table
| Case |
Mode |
Input Data |
Expected Result |
| Simple melting |
Heat needed |
m = 0.25 kg, Lf = 334 kJ/kg, Ti = 0 °C |
83.5 kJ |
| Cold ice correction |
Heat needed |
m = 0.10 kg, Lf = 334 kJ/kg, Ti = -12 °C |
35.9296 kJ |
| Calorimetry test |
Find latent heat |
mice = 0.05 kg, mw = 0.25 kg, Tw = 45 °C, Tf = 24 °C |
Near 337 kJ/kg |
Formula Used
Basic melting heat: Q = m × Lf.
Subzero ice correction: Qwarm = m × cice × (0 − Ti).
Total heat: Qtotal = (Qwarm + m × Lf × f) × loss factor.
Latent heat from supplied heat: Lf = (Qnet − Qwarm) ÷ (m × f).
Mass from heat: m = Qnet ÷ [cice × ΔT + Lf × f].
Calorimetry: Lf = {[(mw × cw + Ccal)(Tw − Tf)] − loss − mice × cice × ΔTice − mice × cw × Tf} ÷ mice.
How to Use This Calculator
Select the calculation mode first. Enter mass in kilograms and heat in kilojoules. Use Celsius for all temperature fields. Keep the default values for specific heat if your problem does not give new values. For laboratory work, choose calorimetry mode. Then enter warm water mass, initial water temperature, final temperature, and calorimeter heat capacity.
Press Calculate. The result appears above the form and below the header. Review each row and the displayed calculation step. Use CSV export for spreadsheets. Use PDF export for a printable record.
Understanding heat of fusion
The heat of fusion of ice is the energy needed to change ice at zero degrees Celsius into water at zero degrees Celsius. The temperature does not rise during this change. Energy breaks the crystal structure instead. This calculator helps you study that hidden energy with several useful modes.
Why corrections matter
A simple textbook problem often uses Q equals m times L. Real lab work may need more detail. Ice can start below zero degrees Celsius. Then it must first warm to the melting point. Warm water may lose heat to the ice, the melted water, and the calorimeter cup. These effects can change the final value. The advanced fields let you include specific heat of ice, specific heat of water, calorimeter heat capacity, and estimated heat loss.
Using results in practice
For design work, the heat needed to melt ice shows the energy load. For a lab report, the calculated latent heat can be compared with the accepted value near 334 kilojoules per kilogram. A large difference may point to wet ice, poor insulation, slow mixing, or thermometer error. The percentage error gives a fast check of data quality.
Good measurement habits
Dry the ice quickly before weighing it. Record the initial ice temperature if it is below zero. Measure hot water mass, starting temperature, and final temperature carefully. Stir gently until the final reading is stable. Avoid splashing, evaporation, and long delays. Enter values using one unit system. This page uses kilograms, kilojoules, and degrees Celsius.
Interpreting outputs
The result box reports the main answer and supporting steps. Heat is also converted to joules and kilocalories. Latent heat is compared with a reference value. The CSV button saves result rows for spreadsheets. The PDF button makes a printable summary. Use the example table to test the calculator before entering your own observations.
Physics meaning
Melting ice is a phase change. It proves that temperature is not the same as heat. Heat is energy transfer. Temperature shows thermal state. During melting, added energy separates molecules while temperature stays fixed. That idea appears in cooling systems, food storage, climate studies, and classroom calorimetry. It also helps students connect equations with careful experimental evidence and uncertainty.
FAQs
What is the heat of fusion of ice?
It is the energy required to change ice at 0 °C into water at 0 °C without changing temperature.
What value is commonly used for ice?
A common reference value is about 334 kJ/kg. Some textbooks use 333.5 kJ/kg or 80 cal/g.
Why does temperature stay constant while ice melts?
Added heat breaks the solid crystal structure. It changes the phase instead of raising the temperature.
When should I use the ice warming correction?
Use it when ice starts below 0 °C. The ice must warm to 0 °C before melting begins.
What does calorimetry mode calculate?
It estimates latent heat from warm water cooling data, ice mass, final temperature, and cup heat capacity.
Why is my lab value far from 334 kJ/kg?
Common causes include wet ice, heat loss, poor stirring, slow reading, evaporation, or inaccurate mass measurement.
Can this calculator find mass of ice melted?
Yes. Select the ice mass mode. Enter supplied heat, latent heat, and any subzero ice temperature correction.
Which units should I enter?
Use kilograms, kilojoules, degrees Celsius, kJ/kg°C, and kJ/°C. The result also shows joules and kilocalories where useful.