Advice needed on how to cool 50 ml liquid from 60 °C to 30 °C

Question

In Short: I'm working on a project where we need to cool 50 ml of liquid from 60 °C to 30 °C or 20 °C. The liquid will be in a 150 ml copper container with an open top.

Due to the lack of skills in the project team (myself & my buddy), we've been looking at the three things we can think of.

1. Peltier style cooling: we have mains power as an option but I'm not sure the time it will take to cool via conduction of roughly 7 cm contact area.

2. CO₂ discharge with some copper pipe submerged in the liquid and a few short discharges of liquid CO₂. We have a few concerns here as to if this would work practically or efficacity with the amount of CO₂ at hand. A further thought on this was a wrapped coil around the container, same concerns as above.

3. Think low-tech, give up on a smart solution and dunk it in an ice bath.

I know this is pretty light on content but I'm struggling to work out what way to focus and trying to be efficient in my use of time and not go down a rabbit hole only to work out the science behind what I was dreaming of would never work out anyway.

Answer 1

Place it in an ice bath with a stirrer.

As Q=m Cp (T2 - T1)

and increasing the container surface area in contact with the ice will improve the rate of heat transfer.

Answer 2

If you're wondering how long it would take in order to cool that liquid.

$$mc_p\frac{dT}{dt} = \frac{1}{\frac{1}h + \frac{\Delta x}k}A(T_C-T) $$

Where:

$m$ is the mass of liquid

$c_p$ is the heat capacity of liquid

$h$ is the convective coefficient of the fluid you're using for cooling (https://www.engineersedge.com/heat_transfer/convective_heat_transfer_coefficients__13378.htm)

$\Delta x$ is the thickness of the copper wall

$k$ is the thermal conductivity of copper, ~400 W/m/K

$A$ is the contact area

$T_C$ is the coolant temperature... which I'm assuming constant. It will be constant if you use a bath of ice or almost constant if you use a lot of coolant, as a thermal reservoir.

But we can pretty much ignore the resistance to mass transfer coming from conductivity, convection is the bottleneck here, so integrating $ mc_p \frac{dT}{dt} = hA(T_C-T) $,

$$ t = \frac{mc_p}{hA}\cdot\ln\left(\frac{T_0-T_C}{T_f-T_C}\right) $$

$T_0$ is the initial temperature, $T_f$ is the final.