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HomeEngineering GuidesAluminum Specific Heat
Material data · calculations · sourcing

Specific Heat Capacity of Aluminum: 6061, 7075, 2024 & 5052

At about 20°C, these four wrought aluminum alloys cluster near 0.88–0.90 J/(g·K). Their difference is small, so specific heat alone rarely decides the best alloy.

The useful answerUse about 897 J/(kg·K) for 6061, 893 for 7075, 875 for 2024, and roughly 880 for 5052 in early calculations. Then verify temperature, temper, product form and source before critical design.
Updated August 2026High-school reading levelTypical values, not lot guarantees
Cut square bar made from 6061 aluminum alloy
6061 aluminum bar. Image: Robert.Baruch / Wikimedia Commons, CC BY-SA 3.0.
8976061 · J/kg·K
8937075 · J/kg·K
8752024 · J/kg·K
≈8805052 · estimated
60-second answer

What should you remember?

Four facts prevent most calculation and buying mistakes.

Same mass

Only a small spread

The room-temperature reference values differ by roughly 2.5% from lowest to highest.

Same volume

Density changes the result

Use volumetric heat capacity, ρcp, when parts occupy the same space.

Heat flow

cp is not conductivity

Specific heat stores energy. Conductivity controls how readily heat moves.

Critical model

Verify the real condition

Attach temperature, temper, product form and source to every design value.

Comparable room-temperature basis

6061, 7075, 2024 and 5052 specific heat compared

Use this table for early calculations—not material acceptance.The first three values come from a NIST-hosted ASM reference at about 20°C. The 5052 value is a common estimate based on similar aluminum alloys, so critical work needs supplier data or testing.
AlloyTypical cp near 20°CJ/(g·K)BtuIT/(lb·°F)Data confidence
6061897 J/(kg·K)0.8970.2142Direct NIST-hosted ASM reference value.
7075893 J/(kg·K)0.8930.2133Direct NIST-hosted ASM reference value.
2024875 J/(kg·K)0.8750.2090Direct NIST-hosted ASM reference value.
5052About 880 J/(kg·K)About 0.880About 0.2102Estimated/common reference; confirm for critical work.

Room-temperature reference: NIST-hosted ASM wrought-aluminum properties. Conversion factor: NIST Guide to the SI. Values are typical references, not guaranteed lot properties.

Why does 7075 sometimes appear as 960 J/(kg·K)?

Kaiser publishes 960 J/(kg·K) for 7075 in a table labeled at 100°C. That value is useful at its stated temperature, but it should not be mixed into a room-temperature ranking. Specific heat changes with temperature, so both 893 near 20°C and 960 at 100°C can be reasonable within their own source conditions.

The decision: If you only need a quick room-temperature estimate, 0.90 J/(g·K) is a reasonable rounded starting point for these alloys. If a few percent matters, use the specific alloy value and record its source. If safety, certification or a wide temperature range matters, use cp(T) data or measure representative material.
Simple calorimeter diagram used to explain heat-capacity measurement
A calorimeter measures heat transferred while a sample changes temperature. Diagram: Li-on / Wikimedia Commons, public domain.
Plain-language definition

What does specific heat capacity mean?

Specific heat capacity, written cp, tells you how much energy is needed to raise one kilogram of material by one degree. A value of 897 J/(kg·K) means that an ideal one-kilogram 6061 sample needs about 897 joules for a one-kelvin temperature rise when that value fits the temperature range.

  • Specific heat cp: energy stored per unit mass and degree.
  • Part heat capacity C = mcp: energy stored by the whole component per degree.
  • Volumetric heat capacity ρcp: energy stored per unit volume and degree.
  • Thermal conductivity k: how readily heat moves through the material.
  • Thermal diffusivity α = k/(ρcp): how quickly a temperature pattern spreads.

Hardness and strength are different again

Heat treatment can change the strength of 6061 or 7075 greatly while changing room-temperature specific heat much less. Do not use cp to predict yield strength, weld strength or corrosion resistance.

No more unit traps

Convert aluminum specific heat without losing a factor of 1,000

Keep the units beside every number. The arithmetic is simple; most errors come from mixing prefixes or mass units.

J/kg·K → J/g·K÷ 1,000

897 J/(kg·K) becomes 0.897 J/(g·K).

J/kg·K → kJ/kg·K÷ 1,000

897 J/(kg·K) becomes 0.897 kJ/(kg·K).

J/kg·K → Btu/lb·°F÷ 4,186.8

897 becomes about 0.2142 BtuIT/(lb·°F).

°C interval → K intervalSame number

A rise of 75°C is a temperature difference of 75 K.

Stored heatQ = m × cp × ΔT

Use for a modest range when one constant cp is accurate enough.

Fixed volumeCvol = ρ × cp

Use when two candidate parts occupy the same envelope instead of having the same mass.

Ideal timet ≈ Q ÷ Pabsorbed

Use net power absorbed by the part—not simply the heater nameplate rating.

Two common 1,000× errors: 0.897 kJ/(kg·K) equals 897 J/(kg·K), not 0.897 J/(kg·K). And 2.70 g/cm³ equals 2,700 kg/m³, not 2.70 kg/m³.
Why websites disagree

Specific heat is a temperature-dependent property

A single cp number is a shortcut for a curve. When temperature rises, aluminum generally needs more heat for each additional degree. A room-temperature value may be good for a small excursion and too simple for an oven, hot-forming cycle or cryogenic system.

Use a constant when the range and risk are modest

For an early estimate around room temperature, Q = mcpΔT is easy to audit. State the chosen cp and run a sensitivity check if published values differ.

Use a curve when the range is wide

For temperature-dependent data, calculate Q = m∫cp(T)dT. If the aluminum melts, add latent heat and use valid solid- and liquid-phase data. A room-temperature constant does not cover a phase change.

Temper and product form still belong in the record

The effect of temper on cp near room temperature is often modest compared with its effect on strength or conductivity. “Modest” does not mean zero. Record the full condition, such as T6, T651, T3, T351 or H32, whenever the model is controlled.

Heat-treatment sequence for producing 6061-T6 aluminum
A typical 6061-T6 heat-treatment sequence. Temper greatly changes strength, while the cp difference near room temperature is usually much smaller. Diagram: Bob Clemintime / Wikimedia Commons, CC BY-SA 4.0.
Alloy-by-alloy interpretation

The cp values are close. The real trade-offs are not.

Choose an alloy from the full mechanical, thermal, corrosion and manufacturing requirement—not one column in a property table.

6061 · 897 J/kg·K

Balanced thermal manufacturing choice

Often the practical starting point for heat sinks, cold plates, housings, machine parts and fixtures.

  • Useful conductivity in common tempers
  • Good availability and machinability
  • Extrudable and generally weldable
7075 · 893 J/kg·K

Strength-led selection

Useful for highly loaded and weight-sensitive parts, but not automatically the best thermal-spreading alloy.

  • High mechanical strength
  • Lower k than 6061-T6 in common high-strength conditions
  • Joining and corrosion strategy need care
2024 · 875 J/kg·K

Aerospace structural trade-offs

Fatigue, damage tolerance, product specification and corrosion protection usually matter more than its small cp difference.

  • Established aerospace applications
  • Condition and product form matter
  • Do not borrow 6061 conductivity data
5052 · about 880 J/kg·K

Forming and corrosion choice

Common for sheet-metal enclosures, tanks, marine components and weldments.

  • Good formability and weldability
  • Strong corrosion resistance
  • The stated cp is an estimate; verify when critical
Interior of an aircraft showing semi-monocoque aluminum-alloy structure
An aluminum-alloy semi-monocoque aircraft structure illustrates an application class; the image does not identify the exact alloy of every member. Image: YSSYguy / Wikimedia Commons, CC BY-SA 4.0.
Worked engineering examples

How much energy is needed to heat the part?

These examples isolate sensible heat stored in the metal. They do not include convection, radiation, fixtures, coolant, contact losses or heater inefficiency.

Step 01

Define the part

Use actual mass—or calculate it from geometry and alloy density.

Step 02

Set the temperature rise

Subtract the starting temperature from the target temperature.

Step 03

Select cp

Match alloy, temperature range, condition and confidence level.

Step 04

Add the real system

Account for losses, fixtures, fluids, interfaces and absorbed power.

10 kg part heated by 75 Kcp usedQ = mcpΔTDifference from 6061
6061897 J/(kg·K)672.75 kJReference
7075893 J/(kg·K)669.75 kJ3.00 kJ lower
2024875 J/(kg·K)656.25 kJ16.50 kJ lower
5052About 880 J/(kg·K)About 660.00 kJAbout 12.75 kJ lower
What this means: The full difference is only about 16.5 kJ, or roughly 2.5%. In a real heating process, part-mass tolerance, fixtures, heat losses and control efficiency may be more important than this alloy-to-alloy cp spread.
From stored energy to cycle time

Estimate warm-up time without overpromising the result

The ideal time is easy to calculate. The useful estimate comes from separating energy stored in the aluminum from every other load and loss in the machine.

Ideal calculation

Start with absorbed power

The 10 kg 6061 example needs 672.75 kJ for a 75 K rise. At a constant 2.0 kW absorbed by the part, the ideal time is 672.75 kJ ÷ 2 kJ/s, or about 336 seconds (5.6 minutes).

Efficiency correction

Nameplate power is not net power

If only 65% of a 2.0 kW heater reaches the part, net absorbed power is about 1.3 kW. The same idealized rise becomes roughly 517 seconds (8.6 minutes), before temperature-dependent losses are added.

System correction

Heat more than the workpiece

Fixtures, clamps, trays, coolant and nearby machine structure may warm during the cycle. Add their heat capacities separately, then account for changing convection, radiation, contact resistance and control behavior.

QuestionInput to collectWhy it changes the answer
What is heated?Part, fixture, fasteners, fluid and insulationEvery heated mass stores energy; omitting a fixture can make the predicted cycle too short.
How is heat delivered?Heater contact, induction, oven, laser or hot fluidThe transfer route changes absorbed power, surface losses and temperature uniformity.
What result matters?Average, surface, core or minimum temperatureA part can reach the right average temperature while still having a damaging hot spot or cold center.
How is the cycle controlled?Sensor position, ramp, soak, duty cycle and feedback delayThe controller may reduce power near the target, so constant-power time is only a first estimate.
A practical design sequence: calculate stored energy, convert heater rating to realistic absorbed power, add fixtures and losses, then compare the estimate with a measured temperature-time curve. If the measured curve is slower or less uniform, update the heat-transfer assumptions before changing the aluminum alloy.
Equal mass is not equal volume

Use ρcp when the part envelope is fixed

Two one-kilogram samples answer a mass-based question. Two one-liter parts answer a packaging question. Because density changes the mass inside one liter, the ranking can change.

AlloyDensity usedVolumetric heat capacity ρcpEnergy for 1 L and 75 K
60612,700 kg/m³2.42 MJ/(m³·K)181.6 kJ
70752,800 kg/m³2.50 MJ/(m³·K)187.5 kJ
20242,770 kg/m³2.42 MJ/(m³·K)181.8 kJ
50522,680 kg/m³About 2.36 MJ/(m³·K)About 176.9 kJ

Which alloy heats fastest?

There is no answer from cp alone. Under the same absorbed power, a lower total heat capacity can raise average temperature faster. But local temperature and uniformity also depend on conductivity, geometry, heat-source footprint, contact, convection and control history.

Simulated temperature distribution across aluminum cooling fins
A simulated temperature distribution across aluminum cooling fins shows why geometry, airflow and conductivity can dominate the final result. Image: Harke / Wikimedia Commons, public domain.
Storage vs spreading

Higher specific heat does not make a better heat sink

A heat sink must move heat from the source into fins and then reject it to air or coolant. Specific heat helps determine transient temperature change, but conductivity and the surrounding heat-transfer path often control the working temperature.

Illustrative diffusivity comparison

Using condition-specific conductivity values with the volumetric heat capacities above gives approximate thermal diffusivities of 69 mm²/s for 6061-T6, 52 for 7075-T6/T651, 50 for 2024-T3/T4, and 59 for 5052-O. These are screening examples because conductivity changes strongly with temper and source.

  • Use cp for energy storage per mass.
  • Use ρcp for energy storage per volume.
  • Use k and the full geometry for steady heat flow.
  • Use α with boundary conditions for transient spreading.
Application decision

Choose the alloy by the job it must do

The best selection combines thermal response with strength, corrosion, forming, welding, machining, product availability and total cost.

Heat sink / cold plate

Start with 6061

Evaluate conductivity, channels, interfaces, flow, flatness, joining and leak testing—not cp alone.

High-load aerospace part

Compare 2024 and 7075

Strength, fatigue, toughness, corrosion, fasteners, service temperature and approved allowables lead the choice.

Formed enclosure / tank

Consider 5052

Formability, weldability, corrosion resistance, gauge, finish, sealing and the H-temper matter most.

Thermal fixture / tooling

Model the whole cycle

Include warm-up energy, temperature uniformity, stiffness, expansion, wear, sensors and control stability.

Fixed-volume thermal buffer

Use ρcp

Density changes how much mass and stored heat fit inside the available package.

Welded thermal assembly

Represent the joint

Local thermal history can change temper, conductivity and mechanical properties around the weld.

Aluminum heat-sink extrusion showing base thickness and cooling fins
An aluminum heat sink shows why material properties and geometry must be evaluated together. Image: Retired electrician / Wikimedia Commons, CC0.
Data quality

When should you test specific heat?

Handbook values are usually enough for concept design and sensitivity work. Testing becomes worthwhile when a small thermal margin, a wide temperature range, qualification, material variation or a contractual requirement makes the source uncertainty important.

ASTM E1269-24

This method uses differential scanning calorimetry to determine specific heat capacity for thermally stable solids and liquids. Its normal operating range is −100°C to 600°C, subject to equipment and specimen limits.

ASTM E1461-13(2022)

The flash method primarily measures thermal diffusivity. Under strict conditions, related cp data may also be determined; conductivity can then be derived from α, density and cp. Each measured or assumed input adds uncertainty.

  • Define alloy, full temper, product form, thickness and lot.
  • Set the temperature range and requested data spacing.
  • Name the method, calibration, sampling and uncertainty.
  • State whether the result is informational or an acceptance requirement.
Aluminum sheet production steps from hot rolling to finished coil
Aluminum sheet production involves several processing stages. Product form and condition should accompany any thermal-property value. Diagram: Sung-Min Wi and Jin A Choi / Wikimedia Commons, CC BY 4.0.
Before FEA or an RFQ

Record the assumption so another engineer can reproduce it

A traceable model is more valuable than a table with many unsupported decimal places.

Material identity

What exactly is being bought?

  • Alloy and complete temper
  • Sheet, plate, extrusion, bar or tube
  • Thickness, coating, cladding or anodizing
  • Welded, bonded or heat-treated condition
Thermal duty

What must the model predict?

  • Start and target temperatures
  • Heat load, duration and duty cycle
  • Warm-up, cool-down and uniformity target
  • Airflow, coolant, contacts and insulation
Evidence level

How certain must the value be?

  • Handbook assumption or supplier value
  • Temperature-dependent curve
  • Lot-specific measured data
  • Method, uncertainty and acceptance rule
Technician welding an aluminum structure during repair
Welding creates a local thermal cycle in which heat input, conductivity, part mass and joint geometry all affect temperature history. U.S. Navy photo by Tyler Caswell, public domain.
Avoid these failures

Seven common mistakes in aluminum heat-capacity calculations

MistakeWhy it failsBetter practice
Using 7075 = 960 at room temperatureThe Kaiser value is labeled at 100°C.Use a room-temperature source or state 100°C clearly.
Calling cp “thermal conductivity”Heat storage and heat flow are different.Model cp, k, density, geometry and boundaries separately.
Mixing J and kJThis creates a 1,000× error.Write units beside every input and result.
Comparing equal-size parts using cp aloneDifferent densities change the part mass.Use ρcp or calculate actual mass.
Assuming highest cp means coolest partConductivity and heat rejection may control temperature.Solve the complete heat path and energy balance.
Using Q = mcpΔT through meltingThe formula omits latent heat and phase-specific properties.Integrate cp(T) and add phase-change enthalpy.
Treating a typical value as guaranteedA handbook number may not represent the purchased lot.Specify method, temperature, sampling and acceptance criteria.
Need a useful material recommendation?

Send the thermal duty—not only the alloy name.

Share the alloy and temper, product form, dimensions or mass, temperature range, heat load, cycle time, environment, joining route, quantity and certification needs. That gives our team enough information to compare manufacturing and thermal risk on the same basis.

Frequently asked questions

Aluminum specific heat FAQ

What is the specific heat capacity of 6061 aluminum?

A NIST-hosted ASM reference lists 897 J/(kg·K), or 0.897 J/(g·K), near 20°C. Use it as a typical reference rather than a guaranteed property for every temper, product form and lot.

What is the specific heat capacity of 7075 aluminum?

A NIST-hosted ASM reference lists 893 J/(kg·K) near 20°C. Kaiser lists 960 J/(kg·K) at 100°C, which explains why both numbers appear online.

What is the specific heat of 2024 aluminum?

A useful typical reference near 20°C is 875 J/(kg·K), equal to 0.875 J/(g·K) or about 0.209 BtuIT/(lb·°F).

What value should I use for 5052 aluminum?

About 880 J/(kg·K) is a common estimate for early work. Because the accessible value is estimated rather than a certified condition-specific result, obtain supplier data or test representative stock when the margin is tight.

Can I use 0.9 J/(g·K) for a quick aluminum calculation?

Yes, for a rough room-temperature estimate across these four alloys. Record the assumption and use alloy-specific or temperature-dependent data when a few percent matters.

Does temper change aluminum specific heat?

It can, because heat treatment changes the material's phase and solute state. Near room temperature, the cp effect is often modest compared with the strength or conductivity change, but critical models should still match the exact condition.

Does higher specific heat make a better heat sink?

No. Higher cp means more energy is needed to change each kilogram by one degree. Heat-sink performance also depends strongly on thermal conductivity, geometry, interfaces, airflow or coolant, and heat-source distribution.

How do I convert J/(kg·K) to Btu/(lb·°F)?

Divide the J/(kg·K) value by 4,186.8 for the International Table Btu. For example, 897 J/(kg·K) is about 0.2142 BtuIT/(lb·°F).

Does a mill certificate guarantee specific heat?

Usually not. Mill certificates commonly report chemistry and mechanical properties required by the product specification. If cp must be guaranteed, add the temperature, method, sampling, uncertainty and acceptance rule to the contract.

Can Q = mcpΔT be used near aluminum melting?

Not by itself. Near and through melting, use temperature-dependent properties and include latent heat. Also confirm the alloy's solidus and liquidus range rather than assuming one pure-metal melting point.

Technical references

Sources behind the engineering guidance

Always verify the current document and the purchased material condition before qualification or contract release.