how do you calculate the volume of cold water needed to cool a hot bath?

Question

for example: if the cold water is at 4 degrees C and you need to cool 200l of water at 80.C down to 40 degrees C (where the hell is the degree symbol on the keyboard!).Is there a formulae for this which would work at any temperature differential and volume?

Answer 1

Sure. The change in heat content of the two liquids before you mix them is equal to the heat content of the mixed liquid. So to write that out mathematically,

m1 c T1 + m2 c T2 = (m1+m2) c Tf

where

m1 = mass of cold water
m2 = mass of warm water
c = specific heat
T1 = temperature of cold water
T2 = temperature of warm water
Tf = final temperature

Another way to think about it is that the heat transfered from the hot water has to equal the heat gained by the cold water. So you could also write out this equation:

m1 c (Tf - T1) = - m2 c (Tf-T2)

Answer 2

When you mix two bodies at different temperatures and when these come to equilibrium at some final temperature, the we use the conservation of energy if no energy is lost to the surroundings. Its form, in the present case, is :
Heat Gained = Heat Lost, gained by one and lost by the other. If you want to use the minus sign with the heat lost, then its form becomes:Heat Gained + Heat Lost = 0.
The formula for the heat gained or lost is m*s* Dt, where m is the mass, s is the specific heat and Dt is the change in temperature.
Hot water: t(1) =80 degrees C, t =40 degrees C( final temperature). Dt = - 40 degrees C, s=1 Cal /g, m = 200.000 g, where we have converted 200 litres in cc's and used the fact that 1 cc of water has a mass of about 1 g.
So : Heat lost by the hot water = 200.000*1*40 =8*10^6 calories.
Let the volume of cold water, with temperature t(2) = 4 degrees C, needed be x litres, then:Here
Dt =t(2) - t = 40 - 4=+36 degrees C.
Heat gained by cold water = 1000 x *1* 36= 36000x calories, where we have again converted x litres into cc's and used the fact that 1 cc of water has a mass of 1 g.
Equating the heat lost by the heat gained, we get:
36000x =8*10^6 which leads to x=8000/36 = 222.2 litres. Answer.
Comment: The answer appears to be reasonable as the final temperature is roughly halfway between the temperatures of cold water and hot water, so the quantity of cold water should not differ fron that of hot water(200 L) too much. You can further refine the argument.

Answer 3

delta Q = m c (Tf - To)

This is the equation you want. Delta Q is the amount of heat required to change the mass m of a substance with a specific heat of c from To to Tf.

The problem you are considering involves TWO substances, the hot water and the cold water. A statement of energy conservation is: The heat lost by the hot water plus the heat gained by the cold water equals zero or

delta Q-hot + delta Q-cold = 0

substitute in mc(Tf-To) for each substance

Mhot Chot (Tf hot - To hot) + Mcold Ccold (Tfcold -To cold) = 0
200kg(4286)(40-80) + Mcold (4286) (40-4) = 0

And then solve for Mcold the mass of the cold water. You can convert to volume after that if you need to.

Answer 4

By definition a big calorie (Cal) is the amount of energy that's needed to bring 1 kg of water at l5 deg. C up one degree C. For simplicity, let's just use Cal as the nergy it takes to bring up 1 Kg (which is also 1 liter) of water up one degree C or the nergy it takes to cool 1 l of water down 1 degree C.

To cool down 200 l of water from 80 C down to 40 C, the water at 80 C needs to lose:

200 x (80-40)= 8000 Cal

That's the same number of calories it will take to bring "v" liters of water at 4 deg. C to 40 deg. C after it mixes with the 80 deg. water:

8000 = (40 - 4) v
8000 = 36 v
v=8000/36 or v = 222.22 liters

The formula therefore can be:

(T1 -T2) v1 = (T2-T3) v2

T1: the temp. of water to be cooled down, T2: final temperature of the mixture
v1: water to be cooled down
T3: temp. of water used as coolant
v2: volume of coolant

Answer 5

Yes there is no doubt in my mind that as some one has already pointed out this is a 'wind up, question you must really be a joy to live with!
The way I work it out is using the 'splosh' method, pour both hot and cold water into the bath then 'splosh it about using my hand until it feels just about right, I then get in. I did at one time try and work out the actual mathematical formula for this but by the time I had sorted it out the water was stone cold, so I gave it up!!
Hope this helps???

Answer 6

I'm not sure BUT the degree symbol ° is easy attained!
Press and hold the ALT button and on the numeric keypad type 0176 and voila... °°°
Hope this helps in a roundabout way! 8¬)

Answer 7

If you add 200 liters of water at 4 C, the temperature will be the average of 4 C and 80C, or 42 C.

final temp = [(200 x 80) + (V x 4) ] (200 + V)
Which is:
V = (16,000 - 200T) / (T - 4)

In this case, you need to add 222.2 liters. Is your bath tub that large?

Answer 8

never mind about all the algebra and all the bullshit a quick remedy that all mums will know if you can get your elbow in the water and keep it there then its the right mixture of hot and cold water

Answer 9

Their is only an answer to a specificate question.As the volume of the water changes so does the sum,Also as the temprature of the water rises and falls so does the sum.Mix both these together and its goes on for ever ,your just trying to wind us up.

Answer 10

When I first clicked on this question I knew what you wanted and know how to do it but to write the math formulia is a nother story, Good luck.