Which of two objects at the same tempreature can cause more intense burns when you touch it: the one with the greater specific heat capacity or the one with the smaller specific heat capacity and why?


4 Answers 4


TL; DR The material with the greater effusivity will be more likely to burn you upon contact.

Analysis of a simplified case

First consider the case where your palm comes into contact constant temperature wall. Often, we can consider your palm as a semi-infinite solid. The requirement for the semi-infinite approximation is that $T(x \rightarrow \infty, t) = T_i$. In other words, the temperature far away from the contact point ($x=0$) at any time $t>0$ is not affected by contact with the solid. This is a reasonable starting point for answering this question.


  1. Both the palm and the object are semi-infinite
  2. Both the palm and the object have constant material properties
  3. Heat transfer is one-dimensional
  4. Neglecting thermal contact resistance

The 1-dimensional Heat Equation can be written as follows:

\begin{equation} \frac{\partial^2T}{\partial x^2} = \frac{1}{\alpha}\frac{\partial T}{\partial t} \end{equation}

where $\alpha = k/(\rho c)$ is the thermal diffusivity, $k$ is the thermal conductivity, $\rho$ is the density, and $c$ is the specific heat capacity. This equation describes the movement of heat within your hand where $x = 0$ is the contact point. The solution to this equation gives us the temperature $T$ as a function of $x$. The only boundary condition we will impose is that of the constant temperature wall $T(x=0)=T_w$.

Because of the semi-infinite approximation, we can use some tricks to solve this equation and get a very nice result:

\begin{equation} \frac{T(x) - T_w}{T_i - T_w} = \mathrm{erf} \left( \frac{x}{2\sqrt{\alpha t}} \right) \end{equation}

where $\mathrm{erf}$ is the error function.

Definition of Thermal Effusivity

Now it turns out that when two semi-infinite surfaces come into contact, they must reach a constant temperature at the contact point, $T_s$ so the above solution holds for this case as well (replace $T_w$ with $T_s$). Furthermore, the energy leaving the warmer solid must equal the energy entering the cooler one. Invoking Fourier's law, we get:

\begin{equation} \frac{-k_A \left( T_s - T_{A,i} \right)}{\sqrt{\pi \alpha_A t}} = \frac{-k_B \left( T_s - T_{B,i} \right)}{\sqrt{\pi \alpha_B t}} \end{equation}

where the subscripts $A$ and $B$ denote thermal properties and initial temperatures of the two solids $A$ and $B$.

If we re-arrange this equation to solve for the surface temperature, we get:

\begin{equation} T_s = \frac{\sqrt{k_A \rho_A c_A}T_{A,i} + \sqrt{k_B \rho_B c_B}T_{B,i}}{\sqrt{k_A \rho_A c_A} + \sqrt{k_B \rho_B c_B}} \end{equation}

where we define $\varepsilon \equiv \sqrt{k \rho c}$. This new property is called the thermal effusivity. It behaves as a weighting factor which 'competes' with the effusivity of the other object. If we use the property in the above equation we get:

\begin{equation} T_s = \frac{\varepsilon_A T_{A,i} + \varepsilon_B T_{B,i}}{\varepsilon_A + \varepsilon_B} = \frac{ \frac{\varepsilon_A}{\varepsilon_B} T_{A,i} + T_{B,i}}{\frac{\varepsilon_A}{\varepsilon_B} + 1} \end{equation}

There are three cases which may help understanding:

\begin{align} &\varepsilon_A \ll \varepsilon_B \quad &\text{then} \quad &\frac{\varepsilon_A}{\varepsilon_B} \rightarrow 0 \quad &\text{and} &\quad T_s \rightarrow T_{B,i} \\ &\varepsilon_A = \varepsilon_B \quad &\text{then} \quad &\frac{\varepsilon_A}{\varepsilon_B} =1 \quad &\text{and} &\quad T_s = \frac{T_{A,i}+T_{B,i}}{2} \\ &\varepsilon_A \gg \varepsilon_B \quad &\text{then} \quad &\frac{\varepsilon_A}{\varepsilon_B} \gg 1 \quad &\text{and} &\quad T_s \rightarrow T_{A,i} \\ \end{align}

Answer So the Temperature will always be skewed towards that of the material with the higher effusivity. You were right a higher heat capacity does mean it's more likely to burn, but it also depends on the thermal conductivity and density.

Will you receive a burn?

Using this analysis we actually have enough information to make a rough estimate of whether contact with a particular material will burn you. Lets consider the human palm. In the absence of scholarly data (I did a little searching, if anyone has some better resources let me know) I have decided to use this nist government page as a reference for the burns associated with a range of skin temperatures. We also need an estimate for the effusivity of human skin. This article reports effusivity of the human palm measured from 6 different subjects. We'll analyze the worst case scenario (e.g. smallest reported effusivity): $\varepsilon \approx 1280 Ws^{1/2}m^{-2}K^{-1}$. And we can pick a few materials which the palm will come into contact with:

| Material                | Density kg/m^3 | Thermal Conductivity W/mK | Heat Capacity J/kgK | effusivity kg/s^2K |
| Oak Wood                | 545            | 0.17                      | 2385                | 470                |
| Asbestos - cement board | 1920           | 0.58                      | 840                 | 967                |
| Concrete                | 2300           | 1.4                       | 880                 | 1683               |
| Copper                  | 8933           | 401                       | 385                 | 37136              |
| Diamond                 | 3500           | 2300                      | 509                 | 64011              |

** Please note, these material properties are not temperature independent. The ones listed here were obtained from a heat transfer text (Incropera and DeWitt) at 300K and Engineering Toolbox at who knows what temperature.

**Also, the materials I do have more data on (Copper and Diamond) have effusivities which don't vary appreciably over the given temperature range.

Assuming the palm is initially at a uniform temperature of $37^oC$, we can plot the contact temperature $T_s$ against the initial material temperature $T_{B,i}$ for the above materials.


I will emphasize that this plot is for illustrative purposes, and is probably not accurate at higher temperatures. For anyone wondering what the 'Arnold Ernst Toht' treatment is, I'll warn you it's not pretty and refer you here. Also, maybe don't use pure copper cookware. In other news you can probably put your face in $55^oC$ asbestos cement board and you will not burn yourself :).


Obviously, the model I described is not great. At all. Though the material properties of the solid shouldn't change that much with temperature, the behaviour of the skin will likely vary substantially over the temperature range. The properties also probably vary a lot with skin depth. Moreover, there's various metabolic reactions going on in the body which mean the flesh is also generating its own heat. I am not an expert in anatomy or physiology so I couldn't tell you how significant those factors are. Also, the semi-infinite approximation is suspect at larger timescales.

  • $\begingroup$ What an excellent, excellent analysis!!!! Bravo. $\endgroup$ Commented Feb 6, 2016 at 12:54
  • $\begingroup$ If this answer doesn't received the bounty, there is something very wrong. $\endgroup$ Commented Feb 10, 2016 at 3:35
  • $\begingroup$ @ChesterMiller Actually, it won't. I didn't quite understand the bounty rules when I made the bounty. I can't award it to my own answer, and when the period expires, half the bounty amount will go to the highest voted answer created after the bounty was started. So probably to Dierk Bormann's answer. $\endgroup$ Commented Feb 10, 2016 at 16:51
  • $\begingroup$ Great answer, which nicely introduces effusivity. There's just a wrong sign in the equality after "invoking Fourier's law". Flux 1 equals minus flux 2 (anyway otherwise you'd get a minus sign in the expression of $T_s$). Also, there's a small mistake in the unit of effusivity in the table. $\endgroup$
    – anderstood
    Commented Nov 12, 2017 at 1:26

Assuming the same mass of the object, the answer is: One with greater specific heat.

As you touch an object heat from the object transfers from the object (warmer object) to your body (colder object). By this process object cools for certain temperature. Greater specific heat means greater heat transferred for the same temperature difference and greater heat/energy means potentially larger injury/burns.

  • 3
    $\begingroup$ Exactly, and also assuming the same thermal conductivity which is a hard requirement for different specific heats. $\endgroup$
    – Alexander
    Commented May 30, 2012 at 20:13
  • $\begingroup$ @Alexander Yes, if you also take a time dimension into consideration, e.g. one touches several objects for the same period of time, which is reasonable assumption, but not provided in OP. $\endgroup$
    – Pygmalion
    Commented May 30, 2012 at 20:15
  • $\begingroup$ Is this a guess? What is the formal definition of the damage? Without any quantitative heat flow rates and an analysis why the mere energy change is harmful, I'm not sure if this is convincing. If I touch a hot object for some seconds, I'd guess that the surface properties or the chemicals involved are more relevant than the overall heat capacity. If there are two objects with different chemistry, which have totally different effects and are uncorrelated regarding the heat capacit, than this answer is only a guideline. $\endgroup$
    – Nikolaj-K
    Commented May 30, 2012 at 20:16

The simple answer to this question is that the specific heat capacity of a hot solid body you touch really doesn't matter too much for whether you burn yourself; what is much more important is its heat conductivity.

The reason for this is that the surface of a hot body with low heat conductivity will rapidly cool down when you touch it (the blood in your body tissue will carry the heat away); afterwards more heat from its inside is only slowly diffusing to the surface and won't hurt you too much. In contrast, a hot body with high heat conductivity will rapidly transport more heat to the surface of contact with your body and thus hurt you more.
In other words, below the surface of a body with low heat conductivity a layer with high temperature gradient and low heat flux is rapidly established, effectively protecting your body from the heat inside, whereas for a body with high heat conductivity the temperature gradient will be low but the heat flux high, leading to more damage.

Moreover, in solid bodies the heat capacity is roughly proportional to the mass density (Dulong–Petit's law) and thus varies only within about an order of magnitude or so anyway, find a table here. Typical heat conductivities, on the other hand, vary by at least three orders of magnitude between good insulators like cork or wood on the low end and good conductors like copper or aluminum on the high end. Still, the bench in my sauna is made of wood, and I really wouldn't recommend you to build yourself one out of aluminum, even if it may look nice and the material is cheaply available in your local store.

Interestingly, the heat capacities per kg of the insulators wood and rubber are larger than those of typical metals, see for instance here. So, what makes hot metals so dangerous to touch is really their high heat conductivity, not their heat capacity.


Consider a spoonful of really hot soup, and say a french fry of similiar temperature - expereince tells you that you burn your mouth 'more' on the soup. As stated, the main difference is the the heat capacity of the soup, that's high because of the high water content.

So yes, a higher heat capacity means you can get more severe burns - assuming beeing burnt badly correlates directly with termal energy flow into your hand (or whatever), and the temperature at which this happens. The with a high capacity object loses temperature slower, so there is a higher $\Delta T$ (temperature difference) for longer, meaning more energy transfer.

A second important characteristic (that is easily checked with a cold piece of metal) is thermal conductivity - the better a thermal conductor your object is, the faster heat flows into the area that you cooled by touching it.

(pygmalions answer including the first comment says the same thing)

  • $\begingroup$ Mhm, the example comparing a liquid with a solid is not 100% convincing. "The [object] with a high capacity loses temperature slower" Is the expression slower here ment in a temporal fashion, unrelated to the last paragraph? If so, do you happen to have a reference for the relation between the heat capacity and the heat flow in time? $\endgroup$
    – Nikolaj-K
    Commented Jun 1, 2012 at 13:12
  • $\begingroup$ at a given point in time you have a $\Delta T$ between the hot and the cold object. the flow of thermal energy is proportional to that $\Delta T$. The change in Temperature (for the hot object) is, given the same loss of energy in one interval of time, smaller for an object with a high thermal capacity. clear now? $\endgroup$
    – mart
    Commented Jun 4, 2012 at 8:11
  • $\begingroup$ So I guess you use a relation like $\Delta H=C\Delta T$, I get that, but now you translated the problem to the justification for "...given the same loss of energy in one interval of time...", i.e. the assumption "In both situations (small $C$ vs. big $C$) the energy flow is the same in a given time interval." $\endgroup$
    – Nikolaj-K
    Commented Jun 4, 2012 at 8:16
  • $\begingroup$ $\Delta H$ for the first interval beeing fix (because the temperature difference between the objects (which I won't call $\Delta T$ anymore) is assumed to be the same at the start of the experioment), the corresponding $\Delta T$ is smaller for a higher $C$ What am I missing about your question? $\endgroup$
    – mart
    Commented Jun 4, 2012 at 8:21
  • $\begingroup$ I'm not certain I understand your last sentence, but my problem is that I don't see where time $t$ enters here. You say "The [object] with a high capacity loses temperature slower", what makes you say that? If you can show that the energy flow is equal in time in both situations, then I understand your $C\Delta T$ argument. $\endgroup$
    – Nikolaj-K
    Commented Jun 4, 2012 at 8:49

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