Submitted by **malahchi** t3_1118kd1
in **askscience**

Let's say that I have water at 255 Kelvin and water at 370 Kelvin. Is there a formula to know which proportion of each I need in order to get water at 313 Kelvin ?

Is there a generalization that include other types of fluids ?

NameUnavailt1_j8dmbk1 wroteAssuming constant heat capacities for the 2 fluids, and no heat loss to the environment we can use conservation of energy to figure this out

Definitions:

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.