Lab 5 Hot and cold - review and XC

In lab 5, the idea was to guess and test an equation for predicting the temperature of a mixture of hot and cold water.

Review

With equal amounts of hot and cold water, most of you predicted that the temperature would be half-way between the temperatures of hot and cold water.

To be more precise, let's define these quantities:

  • $T_H$ is the initial temperature of the hot water (or maybe just warmer than room temperatures) before mixing. To measure this accurately, you must leave a thermometer in the hot water cup until the temperature appears to be unchanging (this usually takes under a minute after putting the thermometer in the cup).
  • $T_C$ is the initial temperature of the cold water (or room temperature water).
  • $T_f$ is the final temperature of the mixture once you've poured one of the cups into the other and the temperature of the thermometers has had a chance to stabilize to the water temp.

The temperature "halfway between" the two temperatures is the same as the average of the hot and cold temperatures. Most of you know how to calculate the average of two numbers like this: $$T_f=\frac{T_C+T_H}{2}.\label{equal}$$ And most of you found in the lab that mixing equal amounts of water resulted in a final temperature of the mixed waters of very nearly half-way between the hot and cold temperatures!

Mix unequal amounts of hot and cold water

Remind yourself about two mathematical facts

  1. You can write a fraction as the product of (1 divided by the denominator) times the numerator: $$\frac34 = 3*\frac14$$ or I could just as well write this as $\frac14*3$.
  2. And if you have two fractions that have the same denominator you can evaluate the sum by adding up the numbers in the numerators like this: $$ \frac 14+\frac34=\frac{1+3}{4}=\frac{4}{4} $$ and 4/4 = 1.

Look again at the expression for the average for equal amounts of water, Eq $\ref{equal}$, I could re-write that as: $$T_f=\frac{T_C+T_H}{2}=\color{blue}\frac12T_C+\frac12T_H.\label{halves}$$ Hmm, with equal amounts of cold and hot water, when I mix them together, the cold water made up half of the final mixture and the hot water made up half of the final mixture.

And the final temperature was half of the cold temperature plus half of the hot temperature.

...Maybe if I have different amounts of hot and cold, I could calculate the fraction of the final mixture that was cold and the fraction of the final mixture that was hot, and then use those fractions in place of 1/2 and 1/2 in Eq. $\ref{halves}$?!

How to calculate these fractions? Let's start with what you measured...

  • $V_C$ is the volume (in milliliters) of the cold water you started with,
  • $V_H$ is the volume of the hot water you started with,
  • $V_f$ is the volume of the water that you end up with, once you have mixed the hot and cold water together
  • ...and $V_f$ can be calculated instead of needing to measure it separately, since... $$V_f=V_C+V_H.$$

With these symbols, the fraction of water in the final mixture that started out cold would be: $$\text{fraction of cold water}=\frac{V_C}{V_f} =\frac{V_C}{V_C+V_H}.$$ So, if we had equal amounts of hot and cold water, $V_H=V_C$ and the fraction above would be $\frac{V_C}{2V_C}=\frac 12$.

and similarly the fraction of the final mixture that started out hot was: $$\text{fraction of hot water }\ =\frac{V_H}{V_f}=\frac{V_H}{V_C+V_H}.$$ and this fraction would also be $\frac 12$ for equal amounts of hot and cold water.

So, let's substitute these fractions instead of $1/2$ and $1/2$ in Eq. $\ref{halves}$:

Final temperature as a "weighted average" of the initial and final temperatures: $$ T_f=\left(\frac{V_C}{V_C+V_H}\right)T_C + \left(\frac{V_H}{V_C+V_H}\right)T_H \label{Tf} $$

Example

In the lab, you tried mixing 100 ml = $V_C$ of cold water with 50 ml = $V_H$ of hot water. The final temperature as predicted by Eq. $\ref{Tf}$ would be: $$\begineq T_f&=\left(\frac{V_C}{V_C+V_H}\right)T_C + \left(\frac{V_H}{V_C+V_H}\right)T_H\\ &=\left(\frac{100}{150}\right)T_C + \left(\frac{50}{150}\right)T_H\\ &=\left(\frac{2}{3}\right)T_C + \left(\frac{1}{3}\right)T_H\\ \endeq $$

Now you try it! (Extra Credit)

  1. Measure out approximately 40 ml of cold water and 100 ml of hot water into two separate styrofoam cups. Record $V_C$ and $V_H$.
  2. Set up an experiment like we did in class...
    1. In logger pro, start your experiment for 400 seconds, measuring 1 sample per second
    2. Place a temperature sensor in each cup and wait while the thermometers settle.
    3. Pour the cold into the hot this time and put both sensors in the cup with the mixture.
    4. Wait until the sensors equilibrate with the mixture
    5. Stop Logger Pro
  3. Examine your graph and your data, and report $T_C$, $T_H$, and $T_f$ of the mixture.
  4. Then use Eq. $\ref{Tf}$ to calculate the expected final temperature of your mixture.
  5. Compare your measured $T_f$ and your expected $T_f$. Were they close? Any ideas about why they might have differed a little bit?