Stirring the Pot: A Simple Question About Heat-Transfer I was preparing hummingbird food. To get the feeders filled and hung quickly, I placed the pot in a sink with a similar or greater volume of cool water in it. I hypothesize that absent any other factors, stirring the water in the pot will decrease the total time until the syrup, sink water and pot reach an equilibrium temperature. I posit that this will increase exposure of all of the liquid in the pot to its walls where heat can be transferred to the sink water, to a smaller degree to the metal sink and even less to the air.
If the water in the sink is also stirred, , more of its mass will be exposed to the pot's outer surface increasing the amount of heat transfered from the simple syrup.
If the waters in both the pot and the sink are stirred, will the time to reach equilibrium be measurably decreased?
Will equilibrium occur more quickly if the two bodies of liquid are stirred in opposite direction (ex: cw + ccw)?
 A: The heat transfer $\dot{Q}$ ($\mathbf{Watt}$) flowing from the hot pot to the cool water is given by Newton's Law of Cooling:
$$\dot{Q}=UA\Delta T(t)$$
where:

*

*$A$ is the contact surface area between pot and cooling water,

*$\Delta T(t)$ is the difference in temperature between the food in the pot and the cooling water surrounding it (both evolving in time $t$),

*$U$ is the overall heat transfer coefficient.

Needless to say, the higher $\dot{Q}$ is, the faster the food cools down.
Note also that:
$$\dot{Q}=0\text{ when }\Delta T(t)=0$$
in other words the cooling of the pot stops when pot and cooling water have reached the same temperature.

As regards the heat transfer coefficient, it can be shown that:

$$\frac{1}{U}=\frac{1}{h_f}+\frac{\theta}{k}+\frac{1}{h_w}$$
where:

*

*$h_f$ is the convection transfer coefficient from food to pot surface,

*$h_w$ is the convection transfer coefficient from pot surface to water,

*$k$ is the thermal conductivity of the pot's material and $\theta$ the pot's wall thickness.

It's well-known that the convection transfer coefficients of liquids increase when the Reynolds number is increased and that this can be achieved by stirring or other forms of agitation.
Obviously, higher values for $h_w/h_f$ also increases $U$.

If the waters in both the pot and the sink are stirred, will the time
to reach equilibrium be measurably decreased?

Yes, in accordance with the above.

Will equilibrium occur more quickly if the two bodies of liquid are
stirred in opposite direction (ex: cw + ccw)?

No, there's no reason to believe that.

In order to calculate the temperature evolution of the food $T_f$ we need first to couple the food temperature and the water temperature, then apply Newton's cooling law.
$$-m_fc_f\frac{\text{d}T_f}{\text{d}t}=m_wc_w\frac{\text{d}T_w}{\text{d}t}=UA\Delta T(t)$$
$$-m_fc_f\text{d}T_f=m_wc_w\text{d}T_w$$
$$-m_fc_f\int_{T_{f,0}}^{T_f}\text{d}T_f=m_wc_w \int_{T_{w,0}}^{T_w}\text{d}T_w$$
$$m_fc_f(T_{f,0}-T_{f})=m_wc_w(T_w-T_{w,0})$$
$$\alpha=\frac{m_fc_f}{m_wc_w}$$
$$\Delta T=T_f-T_w=T_f-\alpha(T_{f,0}-T_f)-T_{w,0}$$
$$\Delta T=(1-\alpha)T_f+\alpha T_{f,0}-T_{w,0}$$
$$-m_fc_f\frac{\text{d}T_f}{\text{d}t}=UA\Big((1-\alpha)T_f+\alpha T_{f,0}-T_{w,0}\Big)$$
$$\beta=\frac{UA}{m_fc_f}$$
$$\frac{\text{d}T_f}{(1-\alpha)T_f+\alpha T_{f,0}-T_{w,0}}=-\beta \text{d}t$$
$$\int_{T_{f,0}}^{T_f}\frac{\text{d}T_f}{(1-\alpha)T_f+\alpha T_{f,0}-T_{w,0}}=-\beta \int_0^t\text{d}t$$
$$\frac{1}{1-\alpha}\ln\Big[\frac{(1-\alpha)T_f+\alpha T_{f,0}-T_{w,0}}{T_{f,0}-T_{w,0}}\big]=-\beta t$$
$$\frac{(1-\alpha)T_f+\alpha T_{f,0}-T_{w,0}}{T_{f,0}-T_{w,0}}=e^{-(1-\alpha)\beta t}$$
$$\boxed{T_f=\frac{1}{1-\alpha}\Big[-\alpha T_{f,0}+T_{w,0}+(T_{f,0}-T_{w,0})e^{-(1-\alpha)\beta t}\Big]}$$
