# B The perfect bath?

1. Jun 25, 2017

### Percy

Is this even Physics, I'm not sure.

Water temperature.
If I have a 100L bathtub. How much boiling water (100C) will I need to get the bath to a nice 38-40C, if the cold water I'm pouring in is 5C.

The practicality of this question is because I built a bath tub outside, but my hot water system is too small to fill it up. So I need to buy a pot big enough to boil and then mix with cold water to enjoy the bath. And pots are expensive, and I dont want to buy one too small then realise it doesn't do the job, or spend more cash on a pot that is over capacity.

Thankyou

2. Jun 25, 2017

### Staff: Mentor

This is a valid physics question. Google for "specific heat examples" and you'll find enough to work it out yourself. We can help you over any hard spots of you get stuck.

3. Jun 26, 2017

### sophiecentaur

Did you consider getting an immersion heater and a long lead? It would be easy / possible to calculate how long a 3kW heater would take to raise the temperature of your bath by the required amount. The cost would be more or less the same whether you heat the water indoors or outdoors.
What other measures do you think you could take to improve the performance?
(Your hot water would not actually be at 100C, btw) Domestic hot water is seldom higher than about 60C (safety issue))

4. Jun 26, 2017

### Percy

100C as i was planning to get a big pot and boil it on a gas ring, then pour it in. the hotwater system isn't big enough, and yes, is set at about 60C.
an immersion heater is something to consider, thanks for the tip.

5. Jun 26, 2017

### olgerm

$\begin{cases} V_{bath}=0.1 \cdot m^3\\ \rho_{water}=980 \cdot kg \cdot m^{-3}\\ t_5=5 \cdot C\\ t_{40}=40 \cdot C\\ t_{100}=100 \cdot C \end{cases}$

$\begin{cases} c \cdot t_5 \cdot m_5+c \cdot t_{100} \cdot m_{100}=c \cdot t_{40} \cdot m_{40} \\ m_5+m_{100}=m_{40} \\ m_{40}=V_{bath} \cdot \rho_{water}\\ m_{100}=V_{100} \cdot \rho_{water} \end{cases}$

$\begin{cases} t_5 \cdot m_5+t_{100} \cdot V_{100} \cdot \rho_{water}=t_{40} \cdot V_{bath} \cdot \rho_{water} \\ m_5+V_{100} \cdot \rho_{water}=V_{bath} \cdot \rho_{water}\\ \end{cases}$

$t_5 \cdot (V_{bath} \cdot \rho_{water}-V_{100} \cdot \rho_{water})+t_{100} \cdot V_{100} \cdot \rho_{water}=t_{40} \cdot V_{bath} \cdot \rho_{water}$

$V_{100}=V_{bath} \cdot \frac{t_{40}-t_5}{t_{100}-t_{5}}\approx 0.037 \cdot m^3 \approx 37 \cdot L$

In these conditions you need 37 liter boiling water. But better buy a bit bigger pot, because water may cool down before you pour it into bathtub.

6. Jun 26, 2017

### Percy

wow! that's some cool Maths.
Thankyou

7. Jun 26, 2017

### CWatters

I think it would be dangerous to put an immersion heating element into a bath.

8. Jun 26, 2017

### nasu

Why? Unless you put "it" and you in the bath at the same time. :)

9. Jun 26, 2017

### Tom.G

Umm, do you expect to carry that 40Kg pot of boiling water to the tub?

10. Jun 29, 2017

### sophiecentaur

Not if you follow the regs. I was rather expecting that a suitable immersion heater could be obtainable for use with a 'flex' / cord.
I bought one for making coffee with - but it was only about 200W!!

11. Jun 29, 2017

### CWatters

How do most people check the temperature of bath water? They stick their hand in it!

Perhaps but wouldn't catch me doing it. Heating elements of all types are notorious for tripping RCD/Earth leakage detectors when they fail.

12. Jun 29, 2017

### sophiecentaur

It would, of course, need to be protected with an RCD and the bath would need to have a permanent Earth. You would be no more at risk than if you got into your bath in your home. If you are really bothered about it then perhaps a separate hot water tank, mounted outside the house and plumbed in, with a dedicated RCD would satisfy you.
Frankly, the bigger risk would be carrying large amounts of scalding hot water from the top of the stove into the garden. The 60 degree figure makes a lot of sense for safety but that wouldn't save your back with an awkward load of water.