I should know this. I’m sure I learned it.
Why do I have to add consideralbly more hot water to my bath to bring it up to the desired temp than cold water to bring it down?
I realize that it’s cooling toward air temperature, but that fast?
Our house temp is 68°F our hot water is 138°F, cold is 58°F and my preferred bath temp is 115°F-118°F. It’s a standard tub, holding about 50 gallons.
( Please, don’t tell me I’m wasting energy, I know.)
I know I had other questions, but I can’t think of them at the moment. I’ll add them when they come back to me.
Thanks.
I would have to suggest the shape of the tub, which is great for getting just about any person into the bath, but not so great for convserving heat.
I need some clarification. I’m assuming:
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You fill up the tub with water at 118 F but it cools down while you are bathing, so you add hot water.
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Assuming #1, you will never need to add cold water.
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Assuming #1 and #2, you will always be adding more hot water than cold.
No, I think the question is that if the person has 50 gallons of water that’s too hot, it takes less cold water to bring it down to the proper temp than it takes of hot water to bring up the temperature if the 50 gallons starts out too cold.
In other words why is it easier to cool down the tub with a little cold water, but it’s hard to warm up the tub with a little hot water.
Exactly.
The tub is cast iron. Not free standing. It’s fully surrounded by heated rooms, including under it. It would have to be warmed up while filling, but that shouldn’t be a factor in adjusting the initial temp.
Well, at least I know the question. I’m absolutely no help with the answer though.
There’s also the fact that by the time you’ve run 50 gallons of water in the tub, be it too hot or too cold, you’ve run a lot of cold water into your hot water heater, and it’s no longer up to temp.
Are you measuring the actual temperature change, or your perception of it?
Also, with a free-standing cast iron tub, you are also heating it up from room temperature with the hot water. Since the tub itself is not pre-heated, some of the heat in the hot water is lost to heating the tub.
You have to define more precisely what you mean my the water being “too hot” or “too cold”. For instance, If you filled the tub with only hot water at 138 F, it would have to cool down by 20 degrees to get to your preferred temp of 118 F. However, when you feel like the water’s too cold, it’s probably around 75-80 F, so you’d have to raise the temperature by about 40 degrees to get it to 118 F. Obviously then you’d use twice as much hot water raising the temperature 40 degrees than using cold water to cool it 20 degrees.
Let’s analyze this. Let’s say we have a tub filled with exactly 100 kg of water and our target temperature is 40°C. First, let’s say that our hypothetical tub is at 45°C. Water has a specific heat of 4186 joule/kg °C. This means that our tub contains 18,837,000 J of energy. Now,if we add 1 kg of the cold water at 15°C, which contains 62,790 J, the total energy will now be 18,899,790 J in 101 kg, which gives a temperature of 44.7°C, a change of .3° C. Another 1 kg of 15°C of water will give 18,962580 J in 102 kg for a temperature of 44.4, and so on.
Now, let’s go the other way. Your hot water is about 12°C higher than your target, while your cold is 30°C lower, so for simplicity, we’ll say our hypothetical hot water is at 50°C. We start off this time below our target temperature, at 35°C. 100 kg of water at this temperature has 14,651,000 J. Adding one kg of the hot water at adds 209,300 J, for a total of 14,860,300 J in 101 kg for a temperature of 35.2 which is less of a change than before. We can go on this way and make complicated graphs of how the temperature changes as we add equal amounts of hot or cold water, but it should be clear at this point that if we are fairly close to the target temperature, your hot water will add less energy per unit than the cold water will, so it takes more to effect a given temperature change because of the greater temperature differential between cold water temp and target temp compared to the hot water temp and the target.
Clear as mud? Me too. I need a bath now to soothe my stressed brain. 
A
Thank you, You’ve given me exactly what I wanted to know. You’ve made the explanation so clear, I feel I should have figured it out for myself. :smack:
Oh, you should add a cup of MgSO4 (Epsom salts) to your bath, it will help relax your muscles.
You’re welcome!
Also, I have NO idea where that “A” at the end came from so nobody wrack your brain trying to figure out what it means. It’s not my new, cryptic signature.
Y’know, I do believe I forgot to convert Celsius to Kelvins. :smack:
In any case, the results will be similar. I’m not doing all that math again at 1:40 in the morning! Someone more ambitious than I is welcome to have at it.
So is the bottom line that your cold water is a lot colder than your hot water is hot? Your preferred temperature is nearly as warm as what the hot water heater will produce, which means when you add hot water, it’s only a tad hotter than what’s in the tub already. But your cold water is way colder than what’s in the tub, so it’ll cool the whole thing off quicker. Is that what’s going on?
Anyway, picunurse, isn’t that a terribly hot bath? Aren’t hot tubs normally in the viscinity of like 105 Fahrenheit or so?
The results will be identical. As long as you’re not involving any phase changes, and the heat capacity of water is constant over the range you’re working with, you can work in degrees Fahrenheit for all it matters and you’ll still get the same results.
That’s a spot-on way of describing it. When you add some volume of water V[sub]1[/sub] at temperature T[sub]1[/sub] to another volume of water V[sub]2[/sub] at temperature T[sub]2[/sub], the resulting temperature is essentially a “weighted average”:
T[sub]final[/sub] = (T[sub]1[/sub] V[sub]1[/sub] + T[sub]2[/sub] V[sub]2[/sub])/(V[sub]1[/sub] + V[sub]2[/sub])
which we can rearrange to read
T[sub]final[/sub] - T[sub]2[/sub] = (T[sub]1[/sub] - T[sub]2[/sub]) V[sub]1[/sub] / (V[sub]1[/sub] + V[sub]2[/sub])
So the change in overall temperature, for a given volume added, is proportional to the difference in temperatures. In other words, water that’s 60 degrees colder is going to make much more of a difference than water that’s 20 degrees hotter.
Absolutely, but the total energy amounts would be higher than what I indicated. The temperatures remain the same.
IMHO, the main reason is that the “too hot” bath is likely to be much closer to the ideal temperature to begin with. Our skin is much more sensitive to heat than luke-warmness, so the difference between “ideal” and “too hot” is smaller than the difference between “ideal” and “too cold”.
Also it’s very unlikely to have to deal with a tub full of 135[sup]o[/sup]F water (20[sup]o[/sup]F above ideal), while it’s quite easy to end up with a tub full of 85[sup]o[/sup]F water (20[sup]o[/sup]F below ideal), either by using too much cold water while filling it up, or by letting it cool down.
Unless you live in Japan and have a heated tub, in which case you can end up with a tub full of scalding-hot water by mistake. I’ve done it, and it does take quite a bit of cold water to bring it down to a berarable temperature. But even then, you have to be very careful because the final transition from “still too hot” to “just right” is very abrupt.