Water becomes cold on mixing energy drink [closed]








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I noticed that the temperature of water decreases noticeably when I mix an energy drink (Glucon-D) in it. The ingredients mentioned on the pack are majorly glucose (52%) and sucrose (45%). I believe that this is due to the inversion of sucrose being an endothermic reaction
$$\ce{C12H22O11 + H2O + \Delta -> \underset{glucose}{C6H12O6} + \underset{fructose}{C6H12O6}}$$
However, I could not find any sources on the internet to validate this. Please confirm/disprove this hypothesis and also explain the correct reason for this.


**EDIT**: After everyone's recommendations of sticking a thermometer in my drink, I did so and came up with the following observations:
\begin{array}{|c|c|c c|c|}
\hline
m\_\text{drink} & V\_\ce{H2O} & T\_i & T\_f & \Delta T\\
\hline
\pu{14.85 g} & \pu{200 ml} & \pu{20.3°C} & \pu{19.5 °C} & \pu{0.7 °C}\\
\hline
\end{array}


Some elementary thermochemistry gives me the result that $\Delta H\_\text{solution}$ for the energy drink is $\approx \pu{39.6 J g-1}$.




Well, the solution enthalpy of sugars is positive.
I found these numbers on the internet


* $\ce{C12H22O11}$ (sugar(sucrose)) : 5.4 kJ/mol
* $\ce{C6H12O6}$ (glucose) : 11 kJ/mol
* $\ce{C6H12O6·H2O}$ (glucose monohydrate) : 19 kJ/mol


So if your "energy drink" is a dry powder (and not a readymade drink in an aluminum can), this *could* explain your observation.


You should however put a thermometer into your experiment, and get us some numbers. As is, my above is just another piece of guesswork.


 2

From the calculations (and from the enthalpy data posted by Karl, reference [here](http://webserver.dmt.upm.es/~isidoro/dat1/Heat%20of%20solution%20data.pdf)), the following can be calculated:


1. Enthalpy of solution of glucose in the mixture, given by
$$\Delta H\_\text{glucose} = \frac{52}{100} \times 14.85 \times \frac 1 {180} \times 11000 = \pu{471.9 J}$$
2. Enthalpy of solution of sucrose in the mixture, given by
$$\Delta H\_\text{sucrose} = \frac{45}{100} \times 14.85 \times \frac 1 {342} \times 5400 = \pu{105.51 J}$$


By calorimetry, we have
$$Q = mc\Delta T = 200 \times 4.2 \times 0.7 = \pu{588 J}$$


Adding the enthalpies of solution should give us a value close to the heat provided by water, and indeed it does!
$$\Delta H\_\text{glucose} + \Delta H\_\text{sucrose} = \begin{array}{|c|} \hline \pu{577 J} \approx \pu{588 J}\\ \hline \end{array} $$


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