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1. Background and The Problem

When you want to prepare a cup of tea, it's best to put the leaves into the water with a good temperature. If the temperature is too low, the tea would not be brewed sufficiently; If the temperature is too high, some of the good nutritions will be gone. However, how to prepare a glass of water with desired temperature? Note that the requirement is not very strict. After all, an error of only a few degree Celsius won't cause a big problem here.

An easy answer: Probe the water with a thermometer. Unfortunately, a thermometer is not always available, and you probably don't want to appear too special or look stupid probing your water each time in front of your friends or colleagues. You may not be experienced enough to tell the rough temperature of the water too. What to do? I thought of a simple method a few years ago. I shouldn't be the first one who discovered this, as the method is derived only with high-school physics. However, I haven't seen anybody talking about this method before, so I write it down here.

2. The Method

This method requires only some hot and some cold water with known temperatures. To prepare some water with a target temperature, just mix some cold and some hot water with a correct ratio. You can calculate the rough ratio with a very simple formula.

Let $T_{cold}$ and $T_{hot}$ be the temperatures of the cold water and the hot water respectively, in Celcius (°C), Kelvin (K), Fahrenheit (°F), or Rankine (°R). To prepare water with a target temperature of $T$ where $T_{cold} < T < T_{hot}$, mix a volume of $V_{cold}$ cold water with a volume of $V_{hot}$ hot water such that:

$\displaystyle \frac{V_{hot}}{V_{cold}} \approx \frac{T - T_{cold}}{T_{hot} - T}$.

If you use ice cubes instead of cold water, use $T_{cold}$ = -70 °C.

Don't worry if you can't really understand the formula: it's less complicated than you think. (You see, good mathematical skills not only make you smarter, they make you healthier here!) There is a very simple graphical interpretation of the formula, and you can perform the calculation with just a ruler.

For example, suppose that we have some cold and hot water at 25 °C and 95 °C respectively, and we want use some water at 80 °C. Let's work out the ratio with the help of a ruler.

1. Decide a scale from the length to the temperature. If you use cm (centimeter) and °C (Celcius), it is often convenient to correspond the 0cm point to 0 °C, and the 10cm point to 100 °C. Thus, 80 °C, say, will correspond to the 8cm point.
2. Mark (visualize or imagine) three points on the ruler that correspond to the temperatures of the cold, the target, and the hot water. In this example, I've marked the points in blue, green, and red respectively as shown in the figure above.
3. The ratio $V_{hot} / V_{cold}$ is equal to the length of cold-target over the length of target-hot. In the figure above, the ratio is thus the length of the violet line over that of the yellow line, which is (8 cm - 2.5 cm) / (9.5 cm - 8 cm) = 5.5/1.5 ≈ 3.67.

Now you know how to make the calculation wisely. But before that, how do you tell the temperatures of the cold and the hot water at the first place? You may even think, "If I have a way to figure out these temperatures easily, then I don't need this method!" Not really. Surprisingly (or not surprisingly for some of you), it is usually not difficult to tell these temperatures.

• Hot water. To brew tea, you need to boil some water. If you use the boiled water within the first few minutes from a normal water kettle, the temperature is about 90 °C - 95 °C. Note that it's usually not about 100 °C (the boiling point), because the temperature of the boiled water drops fast after you have stopped to boil it and have poured some into your glass (probably with room temperature). If you use hot water from a dispenser with temperature indication, then stick to the reading.
• Cold water. If some water has been ready for hours in a bottle or some container, then its temperature is approximately equal to the room temperature. So if you work in an office with comfortable temperature, then the temperature of the water should be around 20 °C - 25 °C. Use the reading from a room thermometer if there is on. (You can refer to the room temperature [en.wikipedia.org] for some details.)
• Ice. If you use ice cubes from a freezer, then use -70 °C as $T_{cold}$. (Read the next section for technical details about this magical temperature.)

After figuring out the ratio, how do you measure the volume of the water required to mix them together? It's simple: Just with your eyes! Or use a spoon if you want. Note that we are only doing approximation calculation here. If the best temperature to brew a type of tea is 80°C, it won't make much difference if you brew it at 82 °C or 78 °C. If you really can't tolerate such an error in tea brewing, skip this article and use a thermometer...

3. Technical Details

3.1. Derivation of the Formula

You can derive the approximation formula above easily by considering the amount of energy required for heating up a constant amount of pure water. You need to make some assumptions to ignore certain negligible effects to obtain it. The assumptions are similar to that used to derive the time required to boil water with a kettle earlier. You should be able to get

$\displaystyle \frac{m_{hot}}{m_{cold}} \approx \frac{T - T_{cold}}{T_{hot} - T}$.

After neglecting the small changes in density of the pure water in the range of 0 °C - 100 °C under the same pressure, we finally get

$\displaystyle \frac{V_{hot}}{V_{cold}} \approx \frac{T - T_{cold}}{T_{hot} - T}$.

If you have good physics sense, you should be able to tell the approximation formula by noticing that the energy required for heating up the pure water is directly proportional to the increase in temperature, and that the specific heat of the pure water does not change much when it is in the liquid form under the same pressure.

3.2. Ice Cubes

People often use ice cubes to cool hot water, so it's desirable to somehow make the formula to work with ice. However, the formula is not directly applicable to ice because, while it takes care of the specific heat capacity, it does not take care of the latent heat of fusion of ice.

Fortunately, we can do a simple trick here to 'emulate' the effect of latent heat. Let's make some assumptions first. We shall consider a typical scenario using ice cubes to cool pure water at 80 °C under standard pressure. The formula derived is still suitable for other common scenarios because the specific heat capacity of the pure water is rather constant. We shall also assume that the ice cube is at 0 °C when we use it. First, let's see what value of $T_{ice}$ we should use to tackle the latent heat in the following formula:

$\displaystyle \frac{m_{hot}}{m_{ice}} \approx \frac{T - T_{ice}}{T_{hot} - T}$,

where $m_{hot}$ and $m_{ice}$ are the masses of the hot water and the ice respectively. We derive this first to avoid neglecting the smaller density of ice as compared with that of water. Think about this: We need to spend some heat energy (which is the latent heat of fusion) to melt some ice at 0 °C to water at 0 °C. How much temperature increase we can gain if we had spent the same amount of heat energy to the same mass of water? It's given by dividing the specific heat of fusion of ice by the specific heat capacity of water. The former is 334 kJ kg^-1, and the latter value is 4.194 kJ kg^-1 K^-1 under our assumptions. So we have 334/4.194 K ≈ 80 K. In other words, we have $T_{ice}$ ≈ -80 °C here, which is 80 K below the freezing point of the pure water.

From the definition of density $\rho = \frac{m}{V}$, we have

$\displaystyle \frac{V_{hot}}{V_{ice}} = \frac{\rho_{ice}}{\rho_{hot}} \cdot \frac{m_{hot}}{m_{ice}} \approx \frac{\rho_{ice}}{\rho_{hot}} \cdot \frac{T - T_{ice}}{T_{hot} - T}$.

Now we have an approximation formula that tackles ice cubes. It has an extra factor $\frac{\rho_{ice}}{\rho_{hot}}$ not found in our original formula. Can we eliminate the factor by choosing another value for $T_{ice}$ so that we can still stick to the same formula? It is possible, if we fix a value for $T$ and can tolerate additional small error introduced when we have a different $T$. Let's denote the new value of $T_{ice}$ as $T'_{ice}$. We have

$\displaystyle \frac{V_{hot}}{V_{ice}} \approx \frac{\rho_{ice}}{\rho_{hot}} \cdot \frac{T - T_{ice}}{T_{hot} - T} \approx \frac{T - T'_{ice}}{T_{hot} - T}$
$\displaystyle \Leftrightarrow \frac{\rho_{ice}}{\rho_{hot}} (T - T_{ice}) \approx T - T'_{ice}$
$\displaystyle \Leftrightarrow T'_{ice} \approx (1 - \frac{\rho_{ice}}{\rho_{hot}})T + \frac{\rho_{ice}}{\rho_{hot}} T_{ice}$

By substituting $\rho_{ice}$ = 0.917 g/cm³ (at 0 °C), $\rho_{hot}$ = 0.97183 g/cm³ (at 80 °C), $T$ = 80 °C, and $T_{ice}$ = -80 °C into the formula above, we have $T'_{ice}$ ≈ -71 °C. Let's just stick to -70 °C, or it would looks odd to use a seemingly accurate -71 °C in our approximation formula. Note that the coefficient of term $T$ in the formula is $1 - \frac{\rho_{ice}}{\rho_{hot}}$, which is quite close to zero since $\rho_{ice} \approx \rho_{hot}$. This means that varying $T$ a bit in our application won't introduce significant error.

I am out of my energy now to do any error analysis of the formula. Anyone? Your feedback about this method is welcomed.