I'm a chemist by training and have experience solving differential equations, though it has been insufficient in solving the problem I have.

Here is the problem:

I brew my own beer and I'm building a cooling apparatus to cool the wort after it has boiled. The cooling apparatus consists of $1/4"$ (d) inch copper pipe (thickness $=0.04"$) that is to be submerged in the wort $(20L)$ with cool water $(10C)$ constantly running through the copper tubing. This is analogous to how a Graham condenser works to cool hot vapours. (In case the dual measurements don't give it away, I'm Canadian. Most everything is metric except our building supplies)

My question:

How long will it take to cool the wort to $37C$? To thermal equilibrium with the cool water $(10C)$ ?


  1. The velocity of the running water fast enough is such that its temperature is constant $dT_a=0$.

  2. The specific heat capacity of water is independent of temperature $dC_s/dT=0$

  3. The thermal conductivity is independent of temperature $dk/dT=0$

  4. The temperature of the water is uniform.

  5. The wort has identical thermal properties to pure water (in reality, the dissolved sugars would alter these values)

Approach: First, an expression for the heat required to cool the wort to a given temperature $T_w$ is derived using specific heat capacity.

$$dq=m×C_s×dT \tag1$$

$$\ \ \ \ q=mC_s(T_w-T_0)$$

$\rm q=heat \ (J)$

$\rm m=mass \ (grams)$

$\rm C_s=specific \ heat \ capacity \ of \ water (J/g)$

$\rm T_0=initial \ wort \ temperature \ (K)$

$\rm T_w= Final \ Wort \ Temperature \ (K)$

Next, using a definition I found for thermal conductivity:

$$dq/dt=k×A/s×dT \tag2$$

where ;

$\rm k=thermal \ conductivity \ of \ copper$

$$=391 W/M^2 \ K$$

$\rm A=surface \ area\ of \ the \ copper pipe$


$$h=12'=3.69m\ \ \ \ \ $$

$$ \ \ \ r=1/8"=0.0032m$$

$$A=0.148m^2 \ \ \ \ \ \ \ \ \ \ \ $$

$\rm s=thickness \ of \ copper\ pipe$

$$= 0.04"=0.001024\ m$$

Integrating Eq. 2 with respect to $dT$ yields Eq. 3.

$$q=k×A/s(T_w(t)-T_a)dt \tag3 $$

This is where I begin to get stuck. So I know the total amount of heat that needs to be transferred to cool the wort to temperature $T_w$ (Eq. 1). I am just a bit confused on how to go about integrating Eq. 3 such that I can determine the time required to cool the wort to the desired temperature. I suspect that I need experimental data to fit a function to $T_w(t)$ as I believe $T_w(t)=Ae^{-zt}$ or something similar. Experimental data would allow me to solve $A$ and $z$.

Am I on the right path? What am I missing that would allow me to determine the time?


2 Answers 2


The analysis of @Puk provides only an extreme lower bound to the amount of time required to cool the wort. This is because the dominant resistance to heat transfer will reside, not within the tube wall, but in the thermal boundary layer situated between the outside of the tube wall and the bulk of the wort fluid. The heat transfer rate (J/s) through this boundary layer will be described by $\dot{Q}=hA(T_{OTW}-T)$, where $T_{OTW}$ is the outside tube wall temperature, T is the wort temperature, h is the heat transfer coefficient, and A is the heat transfer area. In terms of the temperature difference between the water and the wort, this leads to $$\dot{Q}=UA(T_{water}-T)$$where U is the overall heat transfer coefficient (including the tube wall): $$\frac{1}{U}=\frac{s}{k}+\frac{1}{h}$$Puk's analysis can be corrected by replacing k/s in his equations with U.

The magnitude of the outside heat transfer coefficient h is going to depend on the geometry of the bath and cooling coil arrangement, and whether the wort is agitated. If agitation is provided (convective heat transfer), then h will be higher and the cooling time will be less. Otherwise, one must depend on natural convection within the bath, and the cooling time will be longer. Typical values for h in various situations are provided in Transport Phenomena by Bird, et al.


Your Eq. 2 isn't quite right, and there seems to be an abuse of the differential notation going on. You also have the wrong dimensions for $C_s$, the unit should be $\text{J/K/g}$.

The thermal resistance of the pipe wall (assuming $s\ll r$), in analogy with how you would calculate electrical resistance, is $$\frac{s}{kA}.$$ I think you have this right in your equation. But thermal resistance is the proportionality factor between the temperature difference $\Delta T=T_w-T_a$ across the pipe wall, and the heat flow rate $-dq/dt$, so $$\Delta T=-\frac{s}{kA} \frac{dq}{dt}$$ $$ \frac{dq}{dt} = -\frac{kA}{s}\Delta T.$$ This is the same as your Eq. 2, apart from the minus sign (note that $dq/dt$ is negative) and with $dT$ replaced by $\Delta T$ ($dT$ by itself doesn't make sense in that equation). All that remains is to combine this with Eq. 1 and solve:

$$ mC_s \frac{d\Delta T}{dt} = -\frac{Ak}{s}\Delta T.$$ $$\frac{d\Delta T}{dt}+\frac{Ak}{smC_s}\Delta T=0$$ $$\Delta T(t)=(T_0-T_a)\exp\left(\frac{-Akt}{smC_s}\right).$$

The cooling time constant is $$\tau = \frac{smC_s}{Ak}.$$ In practice the wort temperature can be considered to have settled after about $5\tau$, or you can solve the above equation for your target temperature if it is higher than $T_a$.

  • $\begingroup$ Comments split due to character limit. Thank you for a thoughtful reply, I appreciate it. I knew there was some funny business with how I was approaching the integration, but I couldn't quite figure it out. Thank you for the correcting the units as well, I missed that. Rearranging the final equation you derived left me with this expression. Tw(t)=(T0-Ta) exp(-Akt/(60*smCs))+Ta (The 60 is to convert t from seconds to minutes as its easier to understand graphically) $\endgroup$
    – crk05
    Mar 23, 2022 at 8:24
  • $\begingroup$ A visual inspection of a plot of Tw(t) vs t suggests to me that this is a reasonable representation of the problem as there is a smooth decay from Tw(0)=80 with increasing t and an asymptote at 10C. This is quite intuitive and to me suggests this is a good model. From the equation, it would take approximately 35 minutes for the wort to reach 37C. While I don’t have any direct evidence to base this opinion on, this seems like a reasonable amount of time (compared to ~2hrs cooling time from 80C to 37C for 20L of wort in a stainless-steel pot, half submerged in stagnant 10C [initially] water). $\endgroup$
    – crk05
    Mar 23, 2022 at 8:24
  • $\begingroup$ However, as Chet pointed out the rate of cooling is also going to depend on the geometry of the apparatus and whether there is agitation of the wort. I am unsure about the magnitude of the impact of these factors on total cooling time, (i.e. if it leads to significantly longer cooling times) and will discuss this further under Chet’s reply. Once I have this apparatus built I will monitor the temperature over time and report back here to see how close the experimental time is to the predicted and thus if this simple case is a suitable approximation. $\endgroup$
    – crk05
    Mar 23, 2022 at 8:24
  • $\begingroup$ Nevertheless, the model I imagined was definitely a simple/ideal case and I think the expression you derived reflects exactly what I was trying to describe. Thank you for your time and your thoughts. :) $\endgroup$
    – crk05
    Mar 23, 2022 at 8:24
  • $\begingroup$ You are welcome. Chet is right, there will be thermal boundary resistances at the copper-water interfaces in addition to the thermal resistance of the copper itself. This is also partially related to the assumption of uniform temperature in the wort: the heat transfer coefficient will be lower if the uniformity assumption isn't valid. I'm not sure how significant these effects are, they may or may not be important in this case. I think you can still expect the wort temperature to decay roughly exponentially, and your best bet might be to experimentally determine the time constant. Good luck. $\endgroup$
    – Puk
    Mar 23, 2022 at 8:39

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