# General Peltier (lack of) cooling - why?

The physics here are beyond simple.

1ml of water requires 4 watts of cooling to drop 1 degree C.

Thus: 30ml x 4W x 33c = 3.96kW (IE cool 30ml by 33c needs 3.96kW of cooling)

The TEC01-02015 spec sheet shows that at even the meager 12V output at max amps available, shows 33c is the rated dT and the Qc is 100 W. It does not say Qc is per per hour or per minute, so we assume it is constant.

3.96kW divided by 100 W/s = 39.6s OK. Round to 40s.

The TEC is well cooled... the cold side is showing -17c in the aluminum water container attached to the cold side.

WHY can't we simply cool 30cl of water without it taking an hour or more? Yeah, sure, there are losses to the ambiance and to the container holding the water (aluminum). What else are we doing wrong?

• 1 ml of water requires 4.2 watts of cooling to drop 1 degree C in one second as a liquid. Phase changes, e.g. from liquid to ice, need other consideration. You need to use the correct units or else your conclusions will not make sense. – Andrew Morton Jun 28 at 12:29
• 1 mL requires 4.186 joules to change it by 1 degC. Not a good start to your question. – Andy aka Jun 28 at 12:31
• When you say Qc is 100W and it does not say if its per hour or per minute, W are Joules per second so how can you have that per minute or hour? – theguitarfreq Jun 28 at 12:33
• W and J are not interchangeable. – Chu Jun 28 at 12:50
• "W and J are almost interchangeable". No they're not, just as km/h is not interchangeable with km. You might get the same numerical value for each if the integration time is one unit of time measurement but the units are certainly not interchangeable. Tips: Use HTML entity &deg; for the ° symbol and capital 'C' for celcius (lowercase when spelled out, as are all SI units named after a person). – Transistor Jun 28 at 12:54