\[ Q = cm \frac{dU}{dt} \]
So far we have \[ \frac{dU}{dt} = \frac{Q}{cm} \]
For an object, the greater the surface area, the greater the heat exchange \( Q \) with the surroundings.
Similarly, the greater the temperature difference between the object and the surroundings, the greater \( Q \) is.
This leads to Newton's Law of Cooling:
\[ \frac{dU}{dt} = -\frac{hS}{cm} \left( U(t) - u_s \right) \]
\[ \frac{dU}{dt} = - \frac{h S}{cm} (U - u_s) \]
\[ \frac{dU}{dt} = \frac{q}{cm} - \frac{hS}{cm}\left(U - u_s \right) \]
\[ \begin{Bmatrix} \mathrm{rate \, of \, change} \\ \mathrm{of \, heat ~ content } \end{Bmatrix} = 0 \]
\[ J(x) = \begin{Bmatrix} \mathrm{rate \, of \, flow ~ of ~ heat} \\ \mathrm{per ~unit ~ time ~ per \, unit ~ area } \end{Bmatrix} \]
\[ J(x) = -k \frac{dU}{dx} \]
\[ J(x) = -k \frac{dU}{dx} \]