Use Q = mcΔT when the temperature of a substance changes without a change of state. Use Q = mL for a stated mass changing state at its transition temperature. If a process does both, divide it into stages and add the energy for those stages.
These specific heat and latent heat calculations use supplied, constructed data for KCSE revision. They are paper-based models, not instructions for heating substances at home. Values are treated as constant over the specified ranges, pressure is fixed, and stated ideal models neglect energy absorbed by containers and lost to the surroundings.
Read what changes before choosing an equation
| What happens? | Equation | Meaning of the material constant |
|---|---|---|
| Temperature changes within one state | Q = mcΔT | c is specific heat capacity, in J/(kg K) |
| A mass changes state at its transition temperature | Q = mL | L is specific latent heat, in J/kg |
Here Q is transferred energy in joules and m is mass in kilograms. A temperature difference of 1°C equals a difference of 1 K, so the same numerical ΔT can be used for either temperature-difference unit. That does not mean the Celsius and kelvin temperature readings themselves are equal.
The term “specific” means per kilogram. Heat capacity of a whole object is C = mc, measured in J/K. Confusing C with c would introduce an extra or missing mass factor.
For a pure substance changing state at fixed pressure, an ideal heating curve has a plateau at the transition temperature. Energy is still being transferred during that plateau. Temperature staying constant does not mean that Q is zero.
Worked example 1: a temperature rise only
A model calculation considers 0.75 kg of liquid water warming from 22°C to 40°C. Use c = 4,200 J/(kg K). Find the energy gained by the water, assuming no evaporation or other state change.
Temperature rise = 40 − 22 = 18°C, equivalent to 18 K. Therefore Q = 0.75 × 4,200 × 18 = 56,700 J, or 56.7 kJ.
The final reading, 40°C, is not ΔT. Substituting 40 would count an imaginary rise from 0°C. A quick check uses heat capacity: mc = 0.75 × 4,200 = 3,150 J/K. Eighteen equal one-degree increases require 18 × 3,150 J.
This is the energy received by the water. A real source could supply more because a container also warms or energy reaches the surrounding air. Those contributions cannot be calculated without additional data.
Worked example 2: read a complete constructed heating curve
A fictional pure material has mass 0.25 kg, melting temperature 40°C, solid specific heat capacity 800 J/(kg K), liquid specific heat capacity 1,200 J/(kg K), and specific latent heat of fusion 64,000 J/kg. It receives energy at a constant net rate of 80 W.
It begins entirely solid at 10°C and finishes entirely liquid at 70°C. “Net rate” means the rate of energy actually entering the material, after any losses. The supplied constants describe a teaching model, not an identified real chemical.
| Time | Temperature | State at this point |
|---|---|---|
| 0 s | 10°C | Entirely solid |
| 75 s | 40°C | Melting begins |
| 275 s | 40°C | Melting has just finished |
| 387.5 s | 70°C | Entirely liquid |
To sketch the model curve, put time on the horizontal axis and temperature on the vertical axis. Join successive points with straight segments. The middle segment is horizontal. The table supplies all values needed without relying on an image.
Stage A: warm the solid to its melting point
QA = 0.25 × 800 × (40 − 10) = 6,000 J. Time = Q/P = 6,000/80 = 75 s. Use the solid’s c because the material has not yet melted.
Stage B: melt the material at 40°C
QB = mL = 0.25 × 64,000 = 16,000 J. Duration = 16,000/80 = 200 s. This stage starts at 75 s and ends at 275 s, so subtract the time readings rather than treating 275 s as the plateau’s duration.
Within the ideal plateau, solid and liquid coexist. Adding energy changes how much has melted. It does not raise the temperature until the whole stated mass has completed the transition.
Stage C: warm the liquid
QC = 0.25 × 1,200 × (70 − 40) = 9,000 J. Duration = 9,000/80 = 112.5 s. The liquid’s heat capacity differs from the solid’s, so reusing 800 here would be an error.
Total energy = 6,000 + 16,000 + 9,000 = 31,000 J. Total time = 75 + 200 + 112.5 = 387.5 s. Check independently: 80 × 387.5 = 31,000 J.
The first slope is 30/75 = 0.40°C/s; the final slope is 30/112.5 ≈ 0.267°C/s. The liquid warms more slowly under the same net power because its specific heat capacity is larger. A slope comparison needs the same mass and net power, not just two lines on an unrelated graph.
Worked example 3: ice, melting, then liquid water
Calculate the energy for 0.20 kg of ice at −10°C to become water at 20°C. Supply cice = 2,100 J/(kg K), Lfusion = 334,000 J/kg and cwater = 4,200 J/(kg K). Assume melting at 0°C under the stated fixed-pressure model.
- Warm the ice to 0°C: Q1 = 0.20 × 2,100 × [0 − (−10)] = 4,200 J.
- Melt all the ice: Q2 = 0.20 × 334,000 = 66,800 J.
- Warm the resulting water to 20°C: Q3 = 0.20 × 4,200 × 20 = 16,800 J.
Total energy = 87,800 J. The mass remains 0.20 kg throughout because the model has no mass loss. Most energy goes into melting, so ignoring the plateau would drastically underestimate the total.
At a constant net input of 400 W, the ideal time is 87,800/400 = 219.5 s. If instead 400 W is the source’s input power and only 80% reaches the sample, the net power is 320 W. The time becomes 87,800/320 = 274.375 s, approximately 274 s. Always identify which power a question supplies.
Three independent practice tasks
Practice 1: find the temperature increase first
A 0.40 kg solid warms from 18°C to 43°C without changing state. Its specific heat capacity is 900 J/(kg K). Find the energy gain.
Solution: ΔT = 43 − 18 = 25 K. Q = 0.40 × 900 × 25 = 9,000 J. Units cancel as kg × J/(kg K) × K = J. The positive sign represents energy gained by the solid.
Practice 2: a substance already at its melting point
A 0.15 kg solid is already at its melting temperature. Its latent heat of fusion is 180,000 J/kg. How much energy melts it completely, and how long does this take at a constant net 150 W?
Solution: Q = 0.15 × 180,000 = 27,000 J. Time = 27,000/150 = 180 s. No warming term is needed because the solid starts at the transition temperature and the final liquid is at that same temperature.
Practice 3: obtain latent heat from a plateau
A 0.30 kg sample receives a constant net 120 W during a melting plateau lasting 360 s. The whole sample melts. Find its specific latent heat of fusion.
Solution: Energy for melting = Pt = 120 × 360 = 43,200 J. L = Q/m = 43,200/0.30 = 144,000 J/kg. If the plateau interval described only part of the melting process, using the full sample mass would be invalid.
Frequently asked questions
Why not use mcΔT during melting?
For the ideal plateau, ΔT is zero while energy is still required for the transition. The separate latent-heat equation accounts for that energy.
Is latent heat the same for melting and boiling?
No. Fusion refers to solid–liquid change; vaporisation refers to liquid–gas change. Use the value for the actual transition.
Can I use grams with a value in J/kg?
Convert grams to kilograms first. For example, 150 g = 0.15 kg. Mixing these mass units changes an answer by a factor of 1,000.
Does every real heating curve have a perfectly flat plateau?
No. Impurities, changing pressure, measurement conditions and energy losses can complicate real curves. The flat plateau here belongs to an explicitly ideal pure-substance model.
What changes for cooling?
Using ΔT = final minus initial gives negative Q for energy lost by the sample. If asked for the amount released, report its positive magnitude and state the direction.
What if only part of a solid melts?
At the melting point, melted mass = energy used for melting divided by L. Remove any prior warming energy first, and do not let the calculated melted mass exceed the available solid.
Can a plateau duration alone identify a substance?
No. Duration depends on mass, latent heat and net power. You need the other quantities and suitable conditions before extracting a material property.
References and further revision
Standard relations were checked against OpenStax’s temperature change and heat capacity and phase change and latent heat sections. All exercises and curve data here are original. The site’s Physics topical revision resource is login-gated; its member files were not inspected for this lesson.
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