# Loss of temperature of a sphere [closed]

I am trying to get an intuition of whether it theoretically seems possible for a living cell (a yeast for example) to regulate its own temperature.

Let's consider a spherical living cell which radius is $r ≈ 10^{-6}$ meters. The external temperature is $T_{ext}$ °C, while the optimal temperature within the cell is $T_{opt}$ °C and the difference between the two is $\Delta T = T_{opt} - T_{ext}$ °C.

What values of thermal conductivity and of heat production are needed for the cell to maintain its optimal temperature? Are those realistic (to the physicist that you are)?

It is not a homework question but here is my naive try:

If we consider a spherical cell of area $4\pi r^2$, where $r$ is the radius, then the heat production to compensate the heat loss is $4\pi r^2 \cdot \Delta T \cdot K$ (derived form Fourrier's law), where $K$ is the thermal conductivity. Then the ratio of heat production $H$ over the thermal conductivity $K$ (in order to deal with this change of $\Delta T$ °C) is $$\frac{H}{K} = 4 \pi r^2 \Delta T$$.

Estimating the right-hand side

If $r = 10^{-6}$ meters (about the size of a standard cell) and $\Delta T = 2$ °C, then the right-hand side is of the order of $10^{-11}$.

Estimating the left-hand side

The thermal conductivity for water is $k=0.56 \frac{W}{m\cdot K}$(reference-wiki) and the heat production per cell is $34 \cdot 10^{-12}$ W (in a growing colony, according to this paper). So the ratio is of the order of $10^{-11}$ as well!

From those calculations it seems possible to me that individual cells can regulate their internal temperature. Did I get something wrong in my calculations?

## closed as off-topic by Kyle Kanos, ACuriousMind♦, HDE 226868, John Rennie, NeuneckAug 26 '15 at 14:02

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• Is the heat generation uniform within the cell? Or is it concentrated at the center or some other point in the cell? Also, does $T_ext$ refer to the temperature at the cell boundary or the temperature of the surrounding water? – Kyle Arean-Raines Aug 24 '15 at 13:09
• I was picturing an very simplistic model where $T_{ext}$ is the temperature just at the exterior of the cell and the temperature within the cell is constant. But if one could use some slightly more realistic model with for example a radial gradient of temperature within the cell, that would be even better. I am hoping that the answer will be very obvious (Yes/No a cell can('t) regulate it's temperature) so that the slight differences between those models won't be too important. Thanks for your help – Remi.b Aug 24 '15 at 15:13
• I'm not sure I understand what you mean by a constant internal temperature with a different temperature on the exterior. Is there a discontinuity at the cell boundary? You can solve for the radial temperature distribution for uniform heat generation if you have the conductivity within the cell and the temperature at the boundary. If you have the convective coefficient $h$ (dependent on fluid velocity, geometry, and material), you can solve it by specifying the temperature in the bulk (sufficiently far from the cell). See engineeringtoolbox.com/convective-heat-transfer-d_430.html – Kyle Arean-Raines Aug 24 '15 at 16:11

Assuming you fix the temperature at the cell boundary, you can solve for the steady state temperature profile within the cell using (see any elementary heat transfer book)

$$\nabla^2 T = -\frac{\dot q}{k}$$

where $k$ is the thermal conductivity on the inside of the cell and $\dot q$ is the heat generation rate per unit volume.

The temperature only varies radially, so you can reduce this to

$$\frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial T}{\partial r}\right) = -\frac{\dot q}{k}$$

Rearranging terms, integrating once, and applying the boundary condition $\frac{\partial T}{\partial r}|_{r=0} = 0$, you get

$$\frac{\partial T}{\partial r} = -\frac{\dot q}{k}r$$

Integrating again and using the second boundary condition $T(R) = T_{ext}$, where $R$ is the radius of the cell, the temperature profile is

$$T(r) = T_{ext} + \frac{\dot q}{6k}\left(R^2 - r^2\right)$$

You can verify that the maximum temperature occurs at $r = 0$ and is $T_{max} = T_{ext} + \frac{\dot q R^2}{6k}$. The average temperature is

$$\langle T \rangle = T_{ext} + \frac{\dot q R^2}{6k} - \frac{1}{R} \int_0^R dr \frac{\dot q}{6k}r^2 = T_{ext} + \frac{\dot q R^2}{9k}$$

So depending on whether you want $T_{opt} = \langle T\rangle$ or you want $T_{opt} = T_{max}$, you can plug in numbers to get a rough idea.

It's been quite some time since I've done heat transfer problems, so please let me know if I've made mistakes in my work.

The thermal conductivity for water is k=0.56 W/m⋅K,all right,but this is for a lake,for exemple,when the heat transfer is between the air and the cold-water layer.The unit of measurement is W/mK ,not W/(m^2)K. So it's for the thickness ,not for the surface!

If you want to calculate the heat production in the time unit ,H,start from: H=4π(r^2)ΔT α ,where <α>SI = W/(m^2)K.

If the H is 34x10^(-12),then,for your cell, α=1.35 W/(M^2)K. I hope I helped you...