A car is parked outside in the sun. The temperature of the interior of the car reaches 55 Celsius. However a company claims their product, a special spray, can cool down the temperature inside the car down to 25 Celsius in just a few seconds. The spray is a mixture of ten percent alcohol and 90 percent water. Worth 3 marks! please help. I was thinking it has to do with the specific heat capacity of water and thermodynamics :/ oh and btw, this unit is on enthalpy.
What does he question want lol. And is this a test? Btw nice name to go with this post.
lol, thanks? Right, i forgot to include the question :p The question is: "How does the spray work?" Its not a test, just some question from an assignment that i am doing.
@wolfe8
Alright well, you are right about the specific heat. Tell me about the specific heat of water.
water has a high heat capacity about 4.19 j/g.c and so it can absorb more heat from its surroundings
how does it relate to enthalpy though?
Correct. Hmmmm. I'm not sure how you can talk about enthalpy. I suppose you can say water can cause a huge enthalpy change of the surrounding? Not sure. Since it's 3 points worth, do you need 3 points in your answer?
yeah, 3 points the points has to relate to enthalpy or thermodynamics somehow
Hoboy. Well for thermodynamics you can say that heat will move from the warmer air in the car to the cooler spray.
ok, and does the alcohol play a role?
I think alcohol is just to evaporate the solution. Otherwise your car will get wet.
then once the water absorbs the heat, where does the heat go to?
The heat is used to break the bonds of the molecules in the solution I believe. That is also why it evaporates.
So does the hydrogen bonding between the alcohol and water molecules also make the substance absorb more heat in order for their bonds to break and evaporate?
-squints eyes- Not sure. Let's look it up.
i searched and i think evaporation has something to do with the cooling process
Well I think it's what I said. The energy is used to turn liquid into gas basically.
Well, okay. Thank you for your help :)
You're welcome. Not sure I helped much but good luck.
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