# Steady-State Thermal Conduction Calculations

I'm working on a question for class where we need to solve for the heat flux in a system. The problem is stated ad follows:

A $2 \; \mathrm{mm}$ thick glass sheet is being used for a window. The thermal conductivity of glass is $1.7 \; \mathrm{J/(m \cdot K \cdot s)}$. The temperature outside is $-20 \; \mathrm{^\circ C}$ and that inside is $20 \; \mathrm{^\circ C}$. Assuming steady-state is reached, how far into the glass sheet, from the high-temperature side, will the temperature be $+15 \; \mathrm{^\circ C}$.

I understand how to do the problem, I'm just having difficulty with unit correlation.

My first step was to calculate the heat flux. Since the system already reached steady-state I used the following equation: $q=-k \; \mathrm{d}T/\mathrm{d}x$. Then I used the value of $q$ to solve for how far into the glass the temperature will be $15 \; \mathrm{^\circ C}$ : $\Delta x=-k \Delta T/q$

My teacher posted the solutions and my steps were correct just not the values. $k$ is given to us $k=1.7 \; \mathrm{J/m\cdot k\cdot s}$, the temperature in degrees celsius and length in mm. I converted $2\; \mathrm{mm}$ to $0.002 \; \mathrm{m}$. But the solution posted did not correlate the units for temperature. I'm just confused why that is, shouldn't the $20 \; \mathrm{^\circ C}$ be converted to kelvin or vice versa? This way they will cancel out properly in the calculation. Any help is greatly appreciated and I've attached a photo of the solution provided to us to make my question more clear. Thanks

• 1K = 1°C. It's a relatively straightforward conversion. – user121330 Mar 20 '17 at 17:26

## 2 Answers

The temperature difference is the same whether expressed in Kelvin degrees or Centigrade degrees. Your calculation is correct. Since you have not posted the solution which your teacher gave, I don't understand what difficulty you think you have found.

The solution is much simpler than you are making it.

The flux $q$ through the glass is constant (there is nowhere else for it to go to or come from) and the thermal conductivity $k$ is also constant (it does not change with temperature or position in the glass sheet). You can deduce from the equation $q=-k dT / dx$ that the temperature gradient $dT / dx$ is also constant.

This means that the temperature varies linearly with distance from -20C on one side to +20C on the other side. It is a straightforward application of linear algebra to find where the temperature reaches a particular intermediate value. The only other information you need is the thickness of the glass pane (2mm). The values of the thermal conductivity $k$ of the glass and the flux $q$ are both irrelevant.

The Celsius temperature $t$ is defined through the equation (see here)

$$t = T – T_0,\qquad\qquad(1)$$

where $T$ is the thermodynamic temperature and $T_0 = 273.15\,\mathrm{K}$.

Note that, contrary to what many people believe, kelvin and degree Celsius are not two units for the same quantity, but two units for two different quantities, the thermodynamic temperature and the Celsius temperature.

However, from (1) we have $\Delta t = \Delta T$, and the two units are defined so that for any temperature difference,

$$\Delta T/\mathrm{K} = \Delta t/^\circ\mathrm{C}.$$

Since the definition of thermal conductivity involves a temperature difference, this can be expressed either in kelvin or in degree Celsius without changing the numerical value of the thermal conductivity or the equations, that is,

$$\frac{k}{\mathrm{W/(K\,m)}} = \frac{k}{\mathrm{W/(^\circ\mathrm{C}\,m)}}.$$