# What is $k$ in Newtons Law of Cooling? [duplicate]

I am writing the definition of quantities for Newton's Law of Cooling and am not sure what to actually call k$,$ or its units. This is what I have so far

$$\frac{dT}{dt}=-k(T-T_s)$$

Where $\frac{dT}{dt}$ is the rate of cooling measured in degrees Celsius per second $\left[°C\cdot{}s^{-1}\right]$

$T$ is the temperature of the object measured in degrees Celsius $\left[°C\right]$

$T_s$ is the ambient temperature of the objects surrounding measured in degrees Celsius $\left[°C\right]$

$k$ is a constant and has no units? I don't think that's a good enough description

• $T_s$ would have the units $°C$, not $°C·s^{-1}$. – LDC3 May 1 '14 at 4:30
• $k$ in thermodynamics (and quantum mechanics, and some other disciplines touched by either) is Boltzmann's constant. However, in this case, it's a confusingly-named placeholder for a characteristic of the system that may or may not be calculated further on. – Blackbody Blacklight May 1 '14 at 4:33
• Sorry copy and paste error. So it would be correct to name the k in $P=-kA\frac{dT}{dx}$ as the Boltzmann's constant as well? I have it listed as the thermal conductivity coefficient at the moment – user88720 May 1 '14 at 4:35
• @user88720 No, it's not Boltzmann's constant. Nor does it correspond to other variables named $k$ in thermodynamics. – Blackbody Blacklight May 1 '14 at 4:37
• I'm confused. When I look at the Wikipedia pages, it states that both thermal conductivity and the Boltzmann constant are denoted with k... – user88720 May 1 '14 at 4:40

$k$ in thermodynamics (and quantum mechanics, and some other disciplines touched by either) is Boltzmann's constant.
Thermal conductivity (units $\frac{\mathrm W}{\mathrm m^2 \cdot \mathrm K}$) is also named $k$, but that's not what you have there either. Refer to that article for help in deriving what your incomplete equation calls $k$.
• @user88720 Well, the name $x$ tends to emphasize that a variable is supposed to be isolated and calculated. Maybe $c_0$ is a good neutral name. But yeah, such use of $k$ is too confusing. – Blackbody Blacklight May 1 '14 at 4:46