Lead has 207 amu and 125j/kg/c while copper has 63 amu and 376 j/kg/c, why is that? So if energy is stored in the motion of the particles, heavier particles should move slower and therefore wouldn't this means that it require more energy to increase the temperature by 1-degree Celcius? I looked up on google and the main reason it said is that there are more particles, but how? I am taking 1 kg of lead and 1 kg of copper, and from my understanding, their masses are different because the lead particles are heavier instead of more of particles in lead than copper. So please explain why does a heavier element tend to have a low specific heat capacity?
[Physics] Why does a heavier element have a low specific heat capacity
massthermodynamics
Related Solutions
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.
Assumptions:
- Both the palm and the object are semi-infinite
- Both the palm and the object have constant material properties
- Heat transfer is one-dimensional
- 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 :).
Inaccuracies
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.
A better term than "coagulation" is "denaturation". To denature a protein is to unfold it, and this can be done by thermal energy. Initially the egg white proteins are each folded up into little ball-like bundles which limits their interactions, but when they denature the become strands. The charges on these strands then attract to each other and form a tangled mess. Of course, this changes the entropy, but not as much as for simple molecules. That is, we need to look up the "heat of denaturization" for albumin, which is 12.2 J/g (compared to water melting, 334 J/g, so it's fairly small).
That is, denaturing a 60g all albumin egg, should take 60g * 12.2 J/g or 732 J. Given the heat capacity of water as 4200 J/kg K, or 168 J/K for 40 mL of water, this should cause a temperate change of about 4.3C in 40 mL.
So the specific answers are:
The heat of denaturation is small but not completely negligible. Since you start with 100C water, and want to end around 65C, for the small amount of water you mention (40 mL), it's around a 10% error.
Yes, if you're trying to hit it right at 63C, you'll need to account for the heat going into denaturation.
For a temperature controlled system like a sous vide cooker, the amount of energy going into denaturation is negligible compared to energy lost to steam from the bath, etc. (Overall though, this question doesn't quite make sense to me and I wonder whether you're confusing "temperature" with "heat".)
Also, keep in mind that denaturing the albumin does not necessarily mean the egg is cooked sufficiently to be safe to eat.
Finally, I wouldn't trust the 63C number, especially since you're taking the low end of a given range. If you really want to be exact, you'll probably need to go to the source of those numbers. (And don't forget the heat capacity of your container, and also I suspect heat loss will, in fact, be significant here.)
Best Answer
I think you mean to say that they have a different number of particles- 1kg of lead and 1kg of copper surely have the same mass.
There are $~2.91\times10^{24}$ atoms in a kg of lead, and $~9.45\times10^{24}$ atoms in a kg of copper. Heat (more accurately, thermal energy) is "stored" in the particular degrees of freedom for the motion of the atomic "particles"- and since there are more atoms in a kilogram of copper, it's gonna have more atoms with more degrees of freedom. Thus, more capacity for storage.