# Water becomes cold on mixing energy drink [closed]

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 $$\approx \pu{39.6 J g-1}$$.

• Sucrose does not invert upon being dissolved in cold (or room temperature) water. Check out enthalpies of solution. – Ed V Jan 29 at 2:57
• The dissolution of any substance into water is endothermic, because water has to provide some energy to separate the molecules of the solute to be dissolved This is exactly what you have observed. It is a physical phenomena, and no chemical reaction is involved. – Maurice Jan 29 at 16:22
• /The dissolution of any substance into water is endothermic/ That is interesting because if you mix borax powder in water, the water becomes warmer. If you add water to borax powder held in your hand it gets quite hot. Even more true for calcium chloride. @Maurice can you write an answer laying out why some solutes dissolving are endothermic and some are exothermic? – Willk Jan 29 at 16:56
• @Maurice Willk is right. Here's a kinda related answer I wrote some time ago (for precipitations though) which somewhat goes on the same lines. – William R. Ebenezer Jan 29 at 19:00
• @Maurice And you´re really wrong. You cannot dissolve anything without looking closely into the chemistry. Just a few numbers are given here: en.wikipedia.org/wiki/… – Karl Jan 29 at 22:52

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.

• I checked and Glucon-D is a mix of powdered sugars, flavoring, and so on. The solid mix is added to cold water and stirred until it all dissolves. So I think your answer, minus the next to last sentence, is the correct answer. I am ready to upvote if you concur. – Ed V Jan 30 at 2:01
• @EdV I have no idea. ;) "Glucon D" is just a powder? – Karl Jan 30 at 20:24
• Based on what is on the web, it is just a powder mixture. So I am upvoting you answer. And it would be interesting if the OP had a temperature measurement. – Ed V Jan 30 at 20:38
• @EdV Hooray for educated guessing. ;)) – Karl Jan 30 at 20:45

From the calculations (and from the enthalpy data posted by Karl, reference here), 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! $$\Delta H_\text{glucose} + \Delta H_\text{sucrose} = \begin{array}{|c|} \hline \pu{577 J} \approx \pu{588 J}\\ \hline \end{array}$$

• Immediately after posting this comment, I am going to upvote both the question and answer because you listened to the suggestions and comments and then DID THE EXPERIMENT! Bravo! Nice work! It would be so nice if everything went like this. – Ed V Feb 1 at 0:02
• @EdV thank you so much! I really enjoyed doing this experiment: it proved that you don’t need to have access to a lab to enjoy real-world chemistry :) – Aniruddha Deb Feb 1 at 10:34