AskPablo: Water Heater vs. Stove

brass tap.jpg
This week Steve asks: “Which is better, running the tap water for a minute to get hot water or heating cold water on a gas stove?” In order to answer this question I need to state a few assumptions. Let’s say that tap water is 15C (59F) and our water heater is set to 40C (104F), so that will be our desired water temperature. Let’s also assume that the kitchen is on the second floor and the water heater is located in the garage, with 30 feet of pipe in between. By researching on-line I found that a stove burner (natural gas) is less than 50% efficient (think of all the heat that still escapes from underneath the pot). The water heater may have an efficiency of 67% if it is an older model.


Finding the amount of energy required to heat 1 liter of water by 25C is pretty easy. The specific heat of water is 4.1855 J/gK, or Joules required to raise 1g by 1Kelvin. Since 1 liter is equal to 1000g and a change in 25C is equivalent to a change in 25K we get: 4.1855 x (1000g x 25K) = 104,637.5 J, or 105kJ (99.5 Btu). This means that we need to use 210kJ (105kJ/0.50) of Natural Gas (4.5g) on the stove, or 157kJ (105kJ/0.67) of Natural Gas (3.4g) in the water heater.
Unless you are getting your water directly from the water heater it would be much more efficient to heat cold tap water on the stove. This is because a pipe with a half inch interior diameter and a length of 30 feet holds 1.18l, meaning that you have to waste 1.18l of water before you get your 1l of warm water (and the 1.18l of water that follow will quickly lose its heat in the pipe). Of course this assumes that the hot water loses no heat on its way to the tap, which it does. But I barely passed my undergrad Heat Transfer class so I am not about to dust off my old textbooks and dig out my abacus.
Just in case you are interested in how you might proceed, here is a little taste: It is not enough to know the interior area of the pipe, since the pipe has thickness. This means that, as you go further from the center of the pipe there is more and more material to conduct heat away from the water. Some smart fellow (or lass) figured out to use the log mean radius, whatever that is… So, you get the following equation for heat loss from a pipe: (2 x pi x k x L x dT)/ln(ro/ri), where k is the thermal conductivity of the pipe material, L is the length, dT is the temperature difference between the inside and outside of the pipe, ln is the natural log (consult your scientific calculator), ro and ri are the outside and inside radii of the pipe. All this adds up to a massive headache and that’s why I’m not doing it.
See you next week!
Pablo Päster, MBA
Sustainability Engineer
www.AskPablo.org
pablo(dot)paster(at)gmail(dot)com