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My textbook says on conduction:

These 'free' electrons can rapidly carry energy from the hotter to the cooler regions of the metal, so metals are generally good conductors of heat. A metal rod at 20 deg. C feels colder than a piece of wood at 20 deg. C because heat can flow more easily from your hand into the metal.

Wouldn't the metal rod be hotter because more heat is transferred from the hand to the metal?

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Things feel cold when they are sucking heat from you, which metal can do faster than wood. –  ACuriousMind Jul 8 at 1:19
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if they are sucking heat from you, they get hotter and your hand gets colder, so wouldn't the metal feel hotter? –  Oleoleoleole Jul 8 at 1:21
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I don't believe the downvotes are justified, this is a good question, IMO. –  Kyle Kanos Jul 8 at 1:22
    
Adding a school-level reference for what @ACuriousMind mentioned, see the first thermodynamics chapter in Fundamentals of Physics by Resnick, Halliday, Walker. –  New_new_newbie Jul 8 at 5:41
    
As for metals appearing hotter, NO - because even after a very large time, only enough energy would flow so that the temperatures are equalized ($T_{hand} = T_{object}$), it can't suck more heat than that and become hotter than your hand, even locally. You can try holding onto a metal knob/railing in peak winters and see this for yourself. Hold it for 10-15 mins, then leave it for 2-3 seconds and then hold it again. It won't appear significantly colder. –  New_new_newbie Jul 8 at 5:49
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3 Answers 3

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The metal rod will become hotter. Only not very much for a large rod.

The energy will flow from your fingers to the metal until the temperature of the metal reaches the temperature of your fingers. For a large metal object this will never happen for all practical purposes. For a small object, though, it does happen. If you pick up a dime it will initially feel cool, in a minute or two it no longer feels cool.

Note that it takes a minute or so to heat up a dime to the point when it no longer feels cool. For large objects, it would take a very very long time during which the metal is losing energy to the air ... so it may never stop feeling cool.

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When something feels cold, it isn't the temperature that makes it feel cold, it's the rate of heat transfer (heat flux) from your body to the object.

Consider that thermal conduction follows Fourier's Law, $\dot{q} = -k A \Delta T$

We can simplify this equation for our purposes by taking one of the objects to be our body, at body temperature, 37$^\circ$C, and the other test object to always have the same contact area. We can then conclude that the heat flux depends on the thermal conductivity of the object, which is a property of the material it's made of, and its temperature.

$$\dot{q}\sim k~(T - 37)$$

Let's play around with this a bit.

Case 1: Touching the same object, at different temperatures
In this case, $T$ varies. $k$ remains constant. This gives us $\dot{q}\sim(T - 37)$, which is why a dime at 0C feels colder than a dime at 30C

Case 2: Touching a dime and a clay coin, at the same temperature
In this case, $T$ is the same, and the equation reduces to $\dot{q} \sim k$. We know the conductivity of a clay coin is less than the conductivity of a metal coin. That's why we feel the metal coin removing heat from our finger quicker, and hence it feels colder.

Consequently, if you look from the perspective of the coin, it gains heat faster, and hence reaches your body temperature faster. The clay coin will take more time to reach the same temperature.

Since temperature difference is the driving parameter for conduction, both will, given enough time, reach the same temperature.


Now for some slightly more complicated math.

The Fourier equation can be rewritten as $$ \begin{align} \dot{q} &= -k A \Delta T = \frac{\text{d}q}{\text{d}t} \\ \therefore \frac{\text{d}q}{\text{d}t} &= -k A (T - T_{res}) \\ \text{d}q &= m C \text{d}T \\ \therefore m C \frac{\text{d}T}{\text{d}t} &= -k A (T - T_{res}) \\ \therefore \frac{\text{d} T}{T - T_{res}} &= -\frac{k A}{m C} \text{d} t \\ \therefore \int_{T_{init}}^{T}\frac{\text{d} T}{T - T_{res}} &= \int_0^t-\frac{k A}{m C} \text{d} t \\ \therefore \ln(\frac{T - T_{ref}}{T_{init} - T_{ref}}) &= -\frac{k A}{m C} t \\ \therefore \frac{\Delta T}{\Delta T_{init}} &= \exp(-\frac{k A}{m C} t) \\ \therefore \frac{\Delta T}{\Delta T_{init}} &= \exp(-\phi t) \\ \therefore \dot{q} &= -k A \exp(-\phi t) \Delta T_{init} \end{align} $$

As you can see from this picture, the temperature difference falls much quicker if $\phi$ is larger, i.e. $k A / m C$ is larger. If the mass and contact area of both objects under comparison are identical, this implies the larger the $k / C$, the quicker the object reaches equilibrium with the heat reservoir. Here is the same plot, but showing the temperature of the coin initially at 0C with time, when touched to a heat reservoir at 37C.

This next image shows the heat flux plotted against time. As you can see, a smaller $k$ gives a smaller heat flux, but sustained over a longer period of time.


Notes:

$C$ is the specific heat of the material the coin is made of. $k$ is it's thermal conductivity. $m$ is the mass and $A$ is the contact area between the object and the hear reservoir.

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thanks @PranavHosangadi . So if the metal coin and clay coin were the same temperature but above 37 degrees celsius, would the metal coin will feel hotter if I touch it? –  Oleoleoleole Jul 8 at 9:08
    
@oleoleoleole Yes. That's why you take hot drinks in bone china instead of metal –  Pranav Hosangadi Jul 8 at 9:10
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I would say that the textbook is using some pretty vague terms here, although I imagine it's simply to make it easier for you to understand. Different materials have different specific heat capacities, a property which indicates how much heat is required to raise the temperature of the material by 1 deg Celcius per gram of the material. For example, the specific heat of water is 1 calorie per gram per degree Celcius (or, equivalently, 4.184 Joules per gram per degree C -- I use calories here only because it allows me to use simple numbers in my explanation). This means that, if you supplied 50 calories worth of heat, you could raise the temp of 1 gram of water from 25 to 75 C (a change of 50 C), or you could raise the temperature of 50 grams of water from 25 to 26 C, or any other change in which the mass times the change in temperature is 50. Metals generally have very low specific heat values, while more insulating substances like water and wood have higher specific heat values.

Saying something "feels colder" is pretty vague. I would argue that, at the instant you touch either metal or wood they would likely "feel" the same to you. However, if you hold onto say 50 grams of wood and 50 grams of metal, then it will take the metal much less time to reach body temperature. Assuming that your body gives heat to both and the same rate, it takes less energy to raise the temperature of the metal (with a low specific heat) with an equivalent mass of wood (with a high specific heat). You can easily find a table of specific heats for some common materials by just searching the web.

I hope that helps!

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Welcome to Physics.SE . I appreciate that you have added a new aspect in the story - how much heat is required to raise the temperature by a degree (that's what specific heat measures), which certainly would matter in the complete solution of this problem. But one can't sideline thermal conductivity, which is really the reason why metals remove heat faster from the body. You may want to edit your answer, making it more comprehensive. –  New_new_newbie Jul 8 at 7:36
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