Newton’s cooling law says \begin{align*} \overset{\text{flux }\vec{q}}{\overbrace{\left ( -k_{0}\nabla T\right ) }}\cdot \left ( -\vec{n}\right ) & \propto \left ( T_{\Omega }-T_{bath}\right ) \\ \left ( -k_{0}\nabla T\right ) \cdot \left ( -\vec{n}\right ) & =-H\left ( T_{\Omega }-T_{bath}\right ) \end{align*}
Where \(T_{\Omega }\) is the temperature of the surface of body and \(T_{bath}\) is the temperature of the outside.
In the above the proportionality constant \(H>0\,\) always. So \(-H\) is always a negative number.
The above works in all cases. 1D, 2D and 3D and in any configuration. Here is how to use it. Direction of flux vector \(\vec{q}=-k_{0}\nabla T\) is always from hot to cold. We start by drawing \(\vec{q}.\) Now we compare the direction \(\vec{q}\) of to the direction of the reverse of outer normal \(\vec{n}\) to the surface. In other words, we compare the direction of \(\vec{q}\) to the inner normal (not the outer normal), since we are looking at \(-\vec{n}\).
These two vectors are always parallel. They could be in either same direction or in reverse directions.
If direction of \(\vec{q}\) and the inner normal are in the same direction, then the sign on the left is positive. i.e. \(\left ( -k_{0}\nabla T\right ) \cdot \left ( -\vec{n}\right ) \) is a positive quantity (since \(\cos \left ( 0\right ) =1\)).
If the direction of \(\vec{q}\) and inner normal are in the opposite direction, then the sign of \(\left ( -k_{0}\nabla T\right ) \cdot \left ( -\vec{n}\right ) \) is negative (since \(\cos \left ( 180^{0}\right ) =-1\)).
Hence always use \(\left ( -k_{0}\nabla T\right ) \cdot \left ( -\vec{n}\right ) =-H\left ( T_{\Omega }-T_{bath}\right ) \) for \(H>0\). Examples are given below.
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