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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?

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2 Answers 2

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.

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Exactly, and also assuming the same thermal conductivity which is a hard requirement for different specific heats. – Alexander May 30 '12 at 20:13
@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. – Pygmalion May 30 '12 at 20:15
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. – NikolajK May 30 '12 at 20:16

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)

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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? – NikolajK Jun 1 '12 at 13:12
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? – mart Jun 4 '12 at 8:11
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." – NikolajK Jun 4 '12 at 8:16
$\Delta H$ 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? – mart Jun 4 '12 at 8:21
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. – NikolajK Jun 4 '12 at 8:49

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