Stirring the Pot: A Simple Question About Heat-Transfer

The heat transfer $dot{Q}$ ($mathbf{Watt}$) flowing from the hot pot to the cool water is given by Newton's Law of Cooling:

$$dot{Q}=UADelta T(t)$$ 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:

$$dot{Q}=0text{ when }Delta T(t)=0$$

in other words the cooling of the pot stops when pot and cooling water have reached the same temperature.

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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.

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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.

$$-m_fc_ffrac{text{d}T_f}{text{d}t}=m_wc_wfrac{text{d}T_w}{text{d}t}=UADelta T(t)$$ $$-m_fc_ftext{d}T_f=m_wc_wtext{d}T_w$$ $$-m_fc_fint_{T_{f,0}}^{T_f}text{d}T_f=m_wc_w int_{T_{w,0}}^{T_w}text{d}T_w$$ $$m_fc_f(T_{f,0}-T_{f})=m_wc_w(T_w-T_{w,0})$$ $$alpha=frac{m_fc_f}{m_wc_w}$$ $$Delta T=T_f-T_w=T_f-alpha(T_{f,0}-T_f)-T_{w,0}$$ $$Delta T=(1-alpha)T_f+alpha T_{f,0}-T_{w,0}$$ $$-m_fc_ffrac{text{d}T_f}{text{d}t}=UABig((1-alpha)T_f+alpha T_{f,0}-T_{w,0}Big)$$ $$beta=frac{UA}{m_fc_f}$$ $$frac{text{d}T_f}{(1-alpha)T_f+alpha T_{f,0}-T_{w,0}}=-beta text{d}t$$ $$int_{T_{f,0}}^{T_f}frac{text{d}T_f}{(1-alpha)T_f+alpha T_{f,0}-T_{w,0}}=-beta int_0^ttext{d}t$$ $$frac{1}{1-alpha}lnBig[frac{(1-alpha)T_f+alpha T_{f,0}-T_{w,0}}{T_{f,0}-T_{w,0}}big]=-beta t$$ $$frac{(1-alpha)T_f+alpha T_{f,0}-T_{w,0}}{T_{f,0}-T_{w,0}}=e^{-(1-alpha)beta t}$$ $$boxed{T_f=frac{1}{1-alpha}Big[-alpha T_{f,0}+T_{w,0}+(T_{f,0}-T_{w,0})e^{-(1-alpha)beta t}Big]}$$