Is there an 'intuitive' explanation for "Which burns more?"

Question

In helping a friend's son with his grade 10 science homework, I came across a question that essentially asked the following:

"If two objects of equal mass but different specific heat capacities are touched, which will burn more?"

The wording of the question implied that this was meant to be a thought experiment rather than anything calculation-based.

My first reaction was that this question was probably quite a bit more complex than it was made out to be; a quick search online and on this site confirms this, however ideas like conductivity (which I'm familiar with), diffusivity & effusivity (which I'm not) and others are well beyond the student's understanding at the moment. I also felt "burn more" was rather vague.

To address these concerns, I made the following two assumptions:

1. since the question made no reference to time, I took the phrase "burn more" to mean "transfer the most energy by the time equilibrium has been reached"

2. all parameters like mass (mentioned in the question), density, conductivity, area of contact, etc., other than specific heat capacity and time would be equal.

Intuitively I felt the object with the higher specific heat capacity would transfer more energy, but to my slight surprise I found I still couldn't come up with a good explanation, so I told him I'd try to give him one next time we met.

Afterwards, a quick mathematical analysis confirmed my intuition, however it requires elementary calculus or, at a minimum, a rudimentary understanding of rational functions, neither of which the student has.

I saw a few similar questions online and on this site, but most non-mathematical answers gave no explanation. The one that came closest to what I'm looking for said (I'm paraphrasing) that the object with higher specific heat capacity transfers more energy for the same temperature drop than the one with lower specific heat capacity, thus will burn more; however, this seems an incomplete explanation to me as the former will also have a lower overall temperature drop than the latter (i.e. the object with the higher $c$ will have the lower $\Delta T$.)

Is there an intuitive explanation that is more complete than the above one, or is this something I'll only be able to explain convincingly once the student knows a little more math?

Answer 1

I think that effusivity is more relevant in this context. I don't have the original question, but I guess it's along the line of whether you'd get more easily burnt by touching a hot piece of wood vs a hot piece of metal. Since it's grounded, young students can refer to their experience to find the correct answer. In this case, a closer mathematical inspection shows that effusivity, not heat capacity is the quantity to consider.

Intuitively, yes heat capacity will be important as it will increase the total amount of heat transferred. However, the burning sensation still has a temporal component to it, as you can easily handle absorbing a lot of heat over a long period of time. This is why conductivity also comes into play. This cumulative effect can be combined mathematically in a single quantity, effusivity, and double whammy is translated by the product: $$ e = \sqrt{\lambda\rho c} $$

Hope this helps, and tell me if something's not clear.

Answer 2

I think it's wrong to interpret "burn more" as energy transferred by the time equilibrium has been reached, because the energy that the colder object absorbs must be equal to the energy that the hotter object gives out due to conservation of energy. I think the problem here lies in understanding the question.
I can see two interpretations:

1. The problem is asking which object experiences the larger $\Delta T$

Intuitively, you could explain this to someone in a similar way that you have already described in your post: Since Heat Capacity tells you how much energy 1kg of a material needs to increase its temperature by 1 °C, masses and energies being equal, that means that the object with larger $C$ must have lower $\Delta T$. I think it would be even clearer if you would show the equation with the quantities if the kid can understand fractions: $$ C = \frac{ E}{m \Delta T}$$ And you could point out "look this quantity (energy) and this quantity (mass) are the same for both, so if $C$ here is larger, $\Delta T$ here is lower" and viceversa.

2. The problems is asking which objects gives/takes more energy generally for the same $\Delta T$

If this was the case, the question would be wrongly posed in my opinion. However, I think I would explain this intuitively to someone in the same way you described in your post, again showing the equation for $C$ which is justified by the reasoning:
"Heat Capacity tells you how much energy 1kg of a material needs to increase its temperature by 1 °C"

Answer 3

You are right that the term "burn more" is too vague. It could refer to the extent of the burn, how fast the burn occurs, etc.

In any case, the minimum necessary (but not sufficient) condition for a skin burn is the temperature of the object must exceed a certain threshold. Then the severity of a skin burn is determined by the combination of the rate of heat transfer to the skin and the duration of the exposure (contact time). The rate of heat transfer depends on the thermal conductivity and temperature of the object. The product of the heat transfer rate and exposure time is the amount of heat transferred. This is where the heat capacity of the object comes into play.

Generally, the heat capacity of the object only becomes a factor when dealing with contact with small, or low mass parts. A familiar example is aluminum foil. It is possible to touch aluminum foil taken directly from the oven without getting burned, even though the temperature and the thermal conductivity of the foil is very high. This is because the foil is very thin. The combination of the contact area with the foil (about a square centimeter for a finger pad), the thickness of the foil (about 0.016 mm) its density (about 2.7 g/cm$^3$) and its specific heat (about 0.9 kJ/kg C) results in a very low heat capacity, so low that the amount of heat available from the foil is insufficient to cause a skin burn.

For more details about the various factors that go into producing a skin burn, see my answer to the following:

Why does holding a hot object with a cloth make it feel less hot?

Hope this helps.

Answer 4

"If two objects of equal mass but different specific heat capacities are touched, which will burn more?"

I'll add one missing details to the question:

• Let's define "burn" as "thermal watts transferred to a finger over the corresponding contact area when the material is at a specific temperature" or something like Watts/°C of temperature difference between the finger and the material, which is a thermal conductivity.

• Both objects are at the same temperature, which is hot enough to burn a finger

• There is no special magic device to hold both materials at their temperature during the experiment. They are heated, then the heating stops, and the finger is applied instantly. Let's assume the materials don't cool in ambient air.

It is simple to prove that the question cannot be answered without a lot more information. Simply conduct the following experiment:

Grab some aluminium foil from the kitchen. Say, a 30x30cm sheet. Weigh it (should be about 3 grams). Then either cut an aluminium rod so you get a piece of the same weight, or cut another sheet and fold it and bang it with a hammer until you make a solid object out of it. Put them both in the oven, heat to something like 70°C, wait, then ask for a volunteer, open the oven and have the volunteer put one finger on each. You could do the same with a copper block and a copper sponge, or a piece of steel wool and a piece of solid steel.

In all cases, the solid block will definitely feel a lot hotter. If the temperature was higher, it would burn the finger. However, even if the aluminium foil is still inside the oven and we can assume the air around it is still hot and prevents it from quickly cooling to ambient temperature, it won't burn the finger (or at least not that much).

So even with the same material in both cases, the way it is structured will influence how it transfers energy to the finger, because foil, sponge, or steel wool have much lower thermal conductivity than the same solid materials.

So we have to add yet even more missing details to the question:

• The material is a solid block

Now, one might wonder if a "solid block" of styrofoam can exist, since the material is mostly air with a bit of foamed polystyrene holding it. But, oh well, you get the idea. So let's use metals...

Steel: Density 7.8 g/cm3, Heat capacity 500 J/(kg.K), Thermal conductivity 40-60 W/(m.K)

Aluminium: Density 2.7 g/cm3, Heat capacity 921 J/(kg.K), Thermal conductivity 237 W/(m.K)

Copper: Density 8.96 g/cm3, Heat capacity 376 J/(kg.K), Thermal conductivity 401 W/(m.K)

Now we have another problem.

If the pieces of material are smaller than the finger, the one with higher density will be smaller, so it will have lower contact area. So it is less able to transfer heat to the finger. But it will apply the heat to a smaller area, too. So perhaps we should define "burn" as "watts of thermal energy transferred to the finger, per °K of temperature difference and unit area of finger". Or something.

However, if we define "burn" as "damage to the finger" then perhaps we should change it to energy instead of power, so "joules of thermal energy transferred to the finger, per °K of temperature difference and unit area of finger, until the material cools down". Or something. But then we have to know how fast it cools down in ambient air.

If the block of metal is small enough, and its thermal conductivity high enough that we can assume its internal temperature to be uniform, then the case is simple: it can transfer as much energy as it contains due to its heat capacity, so a higher heat capacity material will "burn" more. Whether that would feel hotter or not depends on the contact area, which depends on the size of the block, which means the density, though.

If the block of metal is large, to the point all blocks of all materials contain enough thermal energy to roast the finger, I'd say its heat capacity becomes irrelevant. In this case, what would matter would be the thermal conductivity, or rather effusivity. For this, I don't see how it can be done without writing the differential equation, because you have to consider heat transfer through the material. The initial condition would be a uniform temperature through the material, but as it dumps heat through the finger, a temperature gradient would form...

In other words, the question is nice if it's about prompting "thinking about stuff" but very poor if a one-line answer is expected.

Answer 5

Consider a semi-infinite slab of heat conductive material initially at temperature $T_A$. If the temperature at the surface of the slab is suddenly changed to a lower value $T_S$ for all subsequent times, the heat flux leaving the slab at time t (through the surface of the slab) will be (Transport Phenomena, Bird et al): $$\phi_A=\sqrt{\frac{k_A\rho_A c_A}{\pi t}}(T_A-T_S)$$Similarly, for a slab initially at temperature $T_B$. for which the surface temperature is suddenly changed to a higher value $T_S$ for all subsequent times, the heat flux leaving the slab at time t will be:$$\phi_B=\sqrt{\frac{k_B\rho_B c_B}{\pi t}}(T_S-T_B)$$

If, rather than suddenly changing the surface temperatures of the two slabs individually at time zero, the two slabs were brought into direct contact for all times t > 0, the heat flux leaving slab A would have to equal the heat flux entering slab B: $$\sqrt{\frac{k_A\rho_A c_A}{\pi t}}(T_A-T_S)=\sqrt{\frac{k_B\rho_B c_B}{\pi t}}(T_S-T_B)$$From this we can solve for the contact surface temperature $T_S$: $$T_S=\frac{\sqrt{k_A\rho_A c_A}}{\sqrt{k_A\rho_A c_A}+\sqrt{k_B\rho_B c_B}}T_A+\frac{\sqrt{k_B\rho_B c_B}}{\sqrt{k_A\rho_A c_A}+\sqrt{k_B\rho_B c_B}}T_B$$This relationship is independent of time. It tells us that the higher the value of the quantity $\sqrt{k_A\rho_A c_A}$, the closer the interface temperature will be to that of the higher temperature slab $T_A$.

The above equations also tell us that the heat flux into the colder slab B will be $$\phi_B=\sqrt{\frac{k_B\rho_B c_B}{\pi t}}\frac{\sqrt{k_A\rho_A c_A}}{\sqrt{k_A\rho_A c_A}+\sqrt{k_B\rho_B c_B}}(T_A-T_B)$$This then tells us that the higher the value of the quantity $\sqrt{k_A\rho_A c_A}$, the higher the heat flux will be into the colder slab B.

So in terms of both the temperature at the surface and the heat flux at the surface, increasing $\sqrt{k_A\rho_A c_A}$ causes more "burning." As @lpz correctly points out, this is the key parameter (not just heat capacity).