Heat capacity and thermal conductivity

A pan can spread heat and store heat

Specific heat capacity c measures energy per mass per temperature change under stated conditions, such as approximately constant pressure. Its SI unit is J/(kg·K). Thermal conductivity k relates heat flux to temperature gradient in a conduction model. A material can have a high value of one without a high value of the other.

Two questions about a pan are different: how much energy is needed to bring it to a new temperature, and how readily energy passes through its base. For the first question, define the body mass m and its temperature change ΔT_body. With the appropriate approximately constant c and no phase change, the required heat input is Q = mcΔT_body. Q is an energy amount in joules, not a rate.

Whole-body heat capacity is C = mc, in J/K. Specific heat capacity c is per unit mass, so two pieces of the same material can have the same c and different C. If a supplied c is for approximately constant-pressure heating, use it for that stated condition rather than silently changing the thermodynamic process. Heat names energy transferred because of a temperature difference; the equation accounts for a thermal change, rather than a substance called heat stored inside the body.

A temperature rise of 20°C has the same numerical size as 20 K; an absolute temperature of 20°C is not 20 K. The calculations below use temperature differences. Units, scales and orders of magnitude develops the length/area conversions, while Density and properties relative to mass explains how mass follows from a defined volume.

Include mass and geometry

For a simple temperature interval with approximately constant c and no phase change, Q = mcΔT. Steady one-dimensional conduction through a uniform slab gives heat rate kAΔT/L under suitable boundary conditions. These equations answer different questions.

For conduction, now define a spatial direction x through a slab, from its hot face at x = 0 to its cold face at x = L. Heat flux q_x is energy per time per area crossing a surface, in W/m². In a one-dimensional local Fourier model, q_x = −k(dT/dx). The temperature gradient dT/dx is temperature change per distance, in K/m. Thus k has units W/(m·K). The minus sign makes the flux point down the temperature gradient for a positive k.

Take a uniform slab with constant k, cross-sectional area A normal to x, no internal heat generation and negligible side losses. Once the temperatures are steady, the same heat rate crosses every section. With its actual two faces held at T_hot and T_cold, the profile is linear: dT/dx = (T_cold − T_hot)/L. Therefore the positive hot-to-cold flux is q_x = kΔT_faces/L and the heat rate is Q̇ = A q_x = kAΔT_faces/L, where ΔT_faces = T_hot − T_cold.

Q̇ is read “Q dot”: energy transferred per second. One watt is one joule per second. Check the units: [W/(m·K)] × m² × K/m = W. The rate applies to the entire specified area; flux applies per area. If the steady rate persists for a duration τ, the energy that passes through is Q_through = Q̇τ. That energy need not remain in the slab.

Here the temperatures are the material’s face temperatures, not merely surrounding air temperatures. Surface films, imperfect contacts, radiation and side losses can change a real heat path. A changed boundary condition can change the rate even while the material’s k stays fixed. The first figure shows the direction and gradient for the stated ideal slab.

Compare the same two bodies in two different tests

Use two fictional homogeneous slabs, A and B. Each has face area 0.0100 m², thickness 0.0100 m and density 2000 kg/m³. Volume is AL = 0.000100 m³, so each mass is 0.200 kg. The supplied properties in the table are constant over this teaching interval. They are independent stipulated inputs, not measurements of named materials.

In the temperature-change test, each slab begins uniformly at 20.0°C and ends uniformly at 40.0°C under the stated heat-capacity conditions, without phase change. A needs 0.200 × 1000 × 20.0 = 4000 J, whereas B needs 2000 J. No heating schedule is supplied. These are endpoint energy accounts, not predictions of how fast the interiors warm.

In a separate conduction test, hold one face of each slab at 40.0°C and the other at 20.0°C until steady. The same ΔT_faces = 20.0 K and L give gradient −2000 K/m in each. A then carries flux 400 W/m² and heat rate 4.00 W. B carries flux 4000 W/m² and heat rate 40.0 W. A’s larger c did not give it the larger steady conduction rate.

The second figure uses separate energy and power axes. During 60.0 s of the maintained steady state, A passes 240 J and B passes 2400 J from hot to cold. With no internal generation or other energy changes, energy enters and leaves each slab at equal rates: its temperature profile does not keep rising. Passing energy through a body and raising its temperature are different accounts.

The whole-slab thermal resistance R_th = L/(kA) expresses the same steady model as Q̇ = ΔT_faces/R_th. Its unit is K/W. It is 5.00 K/W for A and 0.500 K/W for B. This includes the stated area; an area-normalized resistance has different units. No interfacial resistance is included in these numbers.

Divide energy by the power that actually changes the body

A time calculation needs a rate with the right physical meaning. In a separate uniform-temperature heating model, suppose the net power delivered as heat to the body is constant, all of that input supports the stated temperature change, and there is no phase change or other changing energy store. The balance is P_net τ = mcΔT_body, so τ = Q/P_net.

For the original 2 kg, c = 500 J/(kg·K), 10 K-rise example, Q = 10000 J. A maintained net heating power of 100 W gives 100 s. If instead 100 W is supplied and a separately stipulated constant 25 W leaves to the surroundings, the net power is only 75 W, giving about 133 s in this uniform-temperature model. A heater rating alone need not be the net power, and real losses may change as temperature rises.

Do not obtain a slab’s actual warm-up time by dividing its uniform-temperature energy by the unrelated steady through-flow rate. A transient temperature field changes with time, and the energy entering a face need not equal the energy leaving the other face then. Thermal diffusivity and response times develops how conduction and heat capacity together affect that evolution under stated boundaries.

Make the measurement answer the intended question

To infer c from a temperature-change experiment, relate the body’s mass and temperature change to the heat actually associated with it. Heat losses, a container or apparatus that also warms, and incomplete temperature uniformity can affect the inference. To infer k in the present slab model, measure the steady heat rate, area, thickness and actual face temperature difference; none of those data alone supplies c. From a property claim to a measurement develops this choice of quantity before equipment.

Both c and k can depend on material state and temperature; k can also depend on direction and structure. If c varies appreciably over the interval, a constant-c calculation needs refinement. A phase change or reaction requires an appropriate additional energy model. Lattice vibrations, electrons and heat transport develops microscopic transport mechanisms, and Interfaces, insulation and thermal management extends the heat path beyond a single ideal slab. The present comparison establishes neither a universal material ranking nor a complete thermal-management design.

Worked examples

Try the reasoning

1. Does heat capacity determine the steady heat rate through a wall?

   Show the answer and reasoning

   Not alone.

   Conductivity, geometry and boundary conditions are needed. In the stipulated steady slab model, c does not appear in Q̇ = kAΔT_faces/L.

2. Can latent heat require a different treatment?

   Show the answer and reasoning

   Yes.

   Phase change adds energy not captured by a simple constant-c rise. The present constant-c examples explicitly exclude phase change; do not infer a melting schedule from their temperature-rise equation.

3. Double slab A’s thickness to 0.0200 m while retaining its material, face area and density. For the same two separate 20 K tests, predict its mass, temperature-change energy and steady conduction rate. Which material properties changed?

   Show the answer and reasoning

   Mass doubles to 0.400 kg, the uniform-temperature energy doubles to 8000 J, and steady conduction rate halves to 2.00 W. The supplied c and k did not change.

   Mass is ρAL, whereas the steady rate is kAΔT_faces/L. This change holds face area fixed; it is not the same as doubling every spatial dimension.

4. An ideal uniform slab carries a measured steady 8.00 W through area 0.00400 m². Thickness is 0.00500 m and the measured hot-to-cold face difference is 10.0 K. Infer k. Does this measure specific heat capacity?

   Show the answer and reasoning

   k = Q̇L/(AΔT_faces) = 1.00 W/(m·K). It does not determine c.

   8.00 × 0.00500/(0.00400 × 10.0) = 1.00. This inversion uses the same no-generation, one-dimensional, constant-k and face-temperature assumptions. No mass/temperature-change heat account was measured.

5. Keep slab B’s thickness, material and face temperatures, but make its face area three times larger. Compare steady heat flux, total heat rate and the uniform-temperature heat input for the same 20 K rise.

   Show the answer and reasoning

   Flux stays 4000 W/m², total steady heat rate becomes 120 W, and the separate uniform-temperature heat input becomes 6000 J.

   Flux kΔT_faces/L is per area. Rate is flux times area. At the same density and thickness, mass and whole-body heat capacity also triple, so the endpoint energy triples. Neither specific c nor k changed.

6. Slab A passes 600 J during 150 s after its fixed-face steady state is established. A learner divides 600 J by its 200 J/K whole-body capacity and claims the slab became uniformly 3 K warmer. Identify the mistaken energy account.

   Show the answer and reasoning

   600/200 = 3 K is the arithmetic for retained heating input, but the stated 600 J passes through. At the stipulated steady state, equal rates enter and leave, so the temperature profile does not rise.

   The rate is 600/150 = 4.00 W, matching the slab calculation. Heat passing the boundary is not automatically a net increase associated with the slab’s temperature. The hot and cold faces remain fixed, and no internal generation or other energy change is present.

What to carry forward

• Q = mcΔT_body gives a temperature-change energy under the stated conditions; C = mc belongs to the whole body.
• Fourier conduction uses a temperature gradient. For the ideal steady slab, flux is kΔT_faces/L and total heat rate is kAΔT_faces/L.
• Joules, watts and watts per square metre are different quantities. A steady through-flow is not automatically net heating power.
• Geometry and boundary conditions belong to the calculation; a transient time requires a matching energy/transport model.