0
$\begingroup$

I have a hot and cold tap which produce water at fixed temperatures $T_C$ and $T_H$ respectively. I'm want to mix the water from the taps to create a volume $V$ of water with a desired temperature $T_D$.

How do I calculate the ratio of hot to cold water needed? What's the theory behind it?

$\endgroup$
  • 2
    $\begingroup$ The final temperature is a weighted average (weighted in terms of the volumes of hot and cold water used) of the hot and cold temperatures. $\endgroup$ – Chet Miller Dec 20 '16 at 12:21
2
$\begingroup$

Energy conservation is the thing. All starting from 1st law of thermodynamics.

No external heat $Q$ or work $W$: $$\Delta U=Q-W=0$$ causes internal energy $U$ to remain unchanged: $$\begin{align} U_{after}&=U_{before}\\ U&=U_{c}+U_{h}\\ mc T&=m_{c}c T_{c}+m_{h}cT_{h}\qquad \leftarrow U=mcT\\ mT&=^*m_{c}T_{c}+m_{h}T_{h}\\ T&=\frac{m_{c}T_{c}+m_{h}T_{h}}{m}\\ T&=\frac{V_c\rho T_{c}+V_h\rho T_{h}}{V\rho}\qquad\leftarrow m=\rho V\\ T&=^{**}\frac{V_c}VT_{c}+\frac{V_h}VT_{h}\\ T&=\frac{V-V_h}VT_{c}+\frac{V_h}VT_{h}\qquad\leftarrow V=^{***}V_c+V_h\\ T&=\left(1-\frac{V_h}V\right)T_{c}+\frac{V_h}VT_{h}\\ \end{align}$$

where $_c$ is cold and $_h$ hot water.

  • At $^*$ we assumed specific heat capacities $c$ constant.
  • At $^{**}$ we assumed densities $\rho$ constant.
  • At $^{***}$ we assumed volumes $V$ constant (incompressible liquid).

For water at normal pressures and house-hold temperatures and small temperature differences all these assumption are good and differ only negligibly.

The final expression gives you $V_h$, which you can put into $V=V_c+V_h$ to find the other one.

$\endgroup$
  • 1
    $\begingroup$ Just for completeness the anwser is $\frac{V_h}{V} = \frac{T-T_c}{T_h-T_c}$ $\endgroup$ – joshlk Dec 22 '16 at 15:02
0
$\begingroup$

The amount of heat in cold water is $Q_C = c m_C T_C$, the amount of heat in hot water is $Q_H = c m_H T_H$, where $c$ is the heat capacity. The heat of the mixture would be $Q_H-Q_C = c (m_C+m_H) T_D$. Combining this with previous formulas you get: $T_D = \frac{m_H}{m_H+m_C}T_H + \frac{m_C}{m_H+m_C}T_C$, which is a weighted average as pointed out in comments. By reshuffling this you can get the ratio between the volumes.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.