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I was running the washing up water this morning, and started to think about why the cold tap isn't hot, and why the water doesn't get hotter the faster it is flowing (if anything, the cold tap gets colder the faster it flows).

From my understanding

$K.E = \frac{mv^2}{2}$

and temperature is directly proportional to kinetic energy.

I know that the $v$ in the above equation is really the mean speed of the particles and therefore some are moving backwards and some moving forwards, it is the speed that is used. But surely the particles of water in the tap are all moving faster, therefore they should all be hotter. Perhaps the particles in the stream are moving at a much higher mean speed than the water is flowing, so the temperature increase is negligible... Am I correct in thinking this? or otherwise, why doesn't the cold tap get hot the more you turn it on?

J_mie6
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3 Answers3

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The water gets colder the longer you run it (in the UK at least) because the water mains pipes buried in the ground are colder than the ones in your house, so sadly this isn't evidence for any fundamental physical effect.

In principle any fluid flowing in a pipe gets hotter because energy is dissipated in viscous flow. You could in principle calculate the energy dissipated using the pressure drop per length of pipe, which is described by the Darcy-Weisbach equation, but this would be a somewhat involved calculation for real pipes/taps and in any case it isn't relevant to the core of your question.

When you relate velocity to temperature you're presumably thinking of the Maxwell-Boltzmann distribution for the temperature dependance of the velocity profile in gases. The trouble is this distribution is arrived at by considering redistribution of energy between gas molecules due to collisions between them. If you simply add a constant velocity to every gas molecule you aren't making any difference to the way the gas molecules collide with each other, because it's only their relative velocities that matter.

Although water is a liquid not a gas the same argument applies. It's the velocities of the water molecule relative to each other that determine the temperature. So just adding a constant velocity to every water molecule makes no difference.

John Rennie
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To greatly simplify John's answer: the temperature gained by friction and velocity is insignificant compared to the current temperature of the liquid.

I expect frictional heating would kick in at higher pressures, but once the water had enough kinetic energy to heat up noticeably on impact with your hands, it would also have enough energy to strip the flesh off your bones, cut through said bones, and punch a hole through the sink. Water jet cutters will easily go through inches of steel and a foot of stone - they aren't overly concerned about heating.

paul
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I totally agree with John Rennie, temperature is related to relative velocity, not bulk velocity.

Still, it may be a good idea, following also paul's argument, to compare what the bulk velocity of the liquid should be for the kinetic energy associated to this velocity to be comparable to the energy in form of heat associated to room temperature. Of course it depends on the liquid, but for water, this velocity should be very roughly of the order of one km/s. Thousands of kms/h. That would be the speed needed for the frictional heating to significantly affect the temperature of the liquid.

It is sometimes difficult to realise how much more energy there is in heat than in other forms of energy, in everyday situations.

Heating 1 gram of water by one degree Celsius is one calorie, about 4 Joules. This is the amount of potential energy corresponding to raising one kilogram by 0.4 meter, or one gram by 400 meters.

A drop of water falling in free fall (in the vacuum) from a height like Mt Everest would only heat up by 20°C upon reaching sea level, if all its kinetic energy (at 400 km/s) is converted to heat.

Alfred
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