You must log in or register to comment.

NameUnavail t1_j8dmbk1 wrote

Assuming constant heat capacities for the 2 fluids, and no heat loss to the environment we can use conservation of energy to figure this out


C: heat capacity (J/kg/K)

T: Temperature (K)

m: Mass (kg)

Energy After mixing: C_Mix × T_Mix × m_mix

Energy before mixing: C_Liquid1 × T_Liquid1 × m_Liquid1 + C_Liquid2 × T_Liquid2 × m_Liquid 2

We can also use the following relations:

m_mix = m_L1 + m_L2

C_Mix = (C_L1 × m_L1 + C_L2 × m_L2)/m_mix

From this we can solve for the mixture temperature, skipping the algebra and getting straight to the result:

T_Mix = (T_L1 × m_L1× C_L1 + T_L2 × m_L2 × C_L2)/(m_L1×C_L1 + m_L2 × C_L2)

In words, we can say that the temperature of the resulting mix is equal to the weighted average of the two input temperatures weighted by their corresponding thermal mass (C×m)

(E: for mixing two portions of the same liquid, the thermal capacities are equal and can be canceled. The result in that case is simply the weighted average of the two Temps, weighted by the mass of each liquid. Since liquids ~ inkompressible you could also weight them by Volume for the same result in this case)

E2: This only applies to (E3: inert) liquids mixing, as u/E_B_Jamisen pointed out, for your given temps you would have ice and hot water mixing, or ice and steam if you're more than a few dozen metres higher than sea level.


N3uroi t1_j8flmtz wrote

Quite good and thorough answer. This only works for mixing liquids with the same (or close enough) chemical compositions though. With two different liquids reactions might take place though, altering the final amount of heat dramatically. The enthaply of mixing might be positive or negative as well. Therefore, mixing two different solutions at the same temperature can consume or produce heat as well.


s0rce t1_j8gvijl wrote

Yah, when you mix methanol and water the solution gets noticeable warmer while acetonitrile and water cools substantially.


Awkward-Motor3287 t1_j8h2yyp wrote

Why does it have to be this complex? Can't it just be calculated like averages are? If mix 1 part 100 degree water and 2 parts 200 degree water, can't I just do the following? (100 + 200 + 200)÷3=166 degrees. Assuming both waters came from the same source of course.


jbhelfrich t1_j8h3h1p wrote

If the two liquids are the same (C_Liquid1 = C_Liquid2) that's exactly what you do get. But the OP asked for a generalized answer for any two fluids.


Cheetahs_never_win t1_j8jhbz3 wrote

No - it's generally a good approximation if at the two states, the fluid's properties are approximately the same.

That's not always true for huge differences in temperature and/or pressure.


NameUnavail t1_j8ha5wt wrote

As I said, so long as it's the same type of liquid you absolutely can. The weighted average for equal portions simply would be the average

But if you have different liquids that doesn't work because different liquids can store different amounts of heat


Scrapheaper t1_j8hn4nh wrote

Only if the heat capacities of the two liquids are identical.

This is actually a very simplistic case since many liquids heat up or cooldown when mixed together due to thermodynamics.


Cheetahs_never_win t1_j8jgch7 wrote

Temperature is only one facet of the total energy state and is simply insufficient on its own to get the job done. I know, we thought the same thing and burned the crap out of ourselves in the shower.

We can't alter the universe's laws for our convenience - we tried, doesn't work. :)

And from the universe's perspective, a temperature scale that's built off a particular kind of matter within a certain range that's convenient for weird ape-people on a specific planet on a specific star is no way to run an entire universe.


E_B_Jamisen t1_j8dnv0m wrote

So its been a while since I have been in college. but if water is at 255 kelvin it is probably ice.

370 kelvin is pretty close to boiling for water.

So my first question is do we need to consider a state transition and the energy required for that. if not then its a matter of ratios (as they are both water). and being honest I would just use excel to figure it out. but it would be something like

(x)(255)+(1-x)(370) = 313

comes out to x = 0.49565217

in terms of engineering - use equal portions and you should have it!!

but once again. if the 255 kelvin water is ice, then you need to calculate the energy of state transition.

good luck!


osorie t1_j8gzmef wrote

Yes, there is a formula to calculate the proportion of water at different temperatures needed to obtain a desired temperature. The formula is known as the "weighted average temperature formula" and it can be expressed as:

(T1w1 + T2w2) / (w1 + w2) = T3

where T1 and T2 are the temperatures of the two water samples, w1 and w2 are their respective weights or amounts, and T3 is the desired temperature.

Using this formula, you can solve for the unknown weight or amount of each water sample needed to obtain the desired temperature.

This formula can also be applied to other fluids as long as their specific heat capacity and density are known.


Radiant-Definition-3 t1_j8ezy9w wrote

yeahh u can easily find the answer by applying specific heat formula which is HEAT(Q)=MASS(M)*CHANGE IN TEMPRATURE(delta T) as specific heat of water is 1cal/g kelvin and for other liquids u can use specific heat of other liquids and find it out. Just equate the heat of both masses.