## Question for anyone...word problem

How many calories are required to raise the temperature of 225g of water from 42C to 75C ?
Thanks to all :cry:

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Cp of water (liquid form) = 75.44 J/mol*K

Change in temperature = 33 K

Heated needed for [b]1 mole[/b] of H2O = Cp * (change in temperature)

you have 225g / (16 + 1 + 1 g/mol) = 12.5 moles

Multiply everything together, convert from Joules to calories

4.186 J = 1 calorie

Remember the conversion in between Joule and Calorie is just approximation. Don't take it for large numbers (10000s)

Approximation is quasi-correct. When your talking about a difference of 4.186 J or 4.144 J, it doesn't matter. One-hundreth of a Joule is such a tiny ammount of energy, when would it matter? Especially since most of this is for homework. Just my two cents. Stick with SI units like Joules, and you can't go wrong :D

By the way, the farther I got into chemistry, science, engineering etc, the more I realized [b]everything[/b] is an approximation. I thought Boyle's gas law was it back in the day. PV=nRT works for very few gases in a very small temperature range. There are litterally hundreds of different approximations, all with better results for certain types of gases in certain temperature regimes. As long as you are within say "5%" of the mark, your answer works in my book :wink:

welcome to the real world lol

My father asked me if I'm being an engineer or a doctor. I answered anyway these have something to do with structuring. And you are dealing with people's lives. That's the annoying real world (but you have to suffer if you want to leave) :x

I chose engineering. I can't stand the sight of blood, nor can I tolerate needles. I would have "tossed my cookies" too many times, before I even got out of med school. Technology is much more fun!

Nah, choosing neither is not a good way. Now there are too many people fighting for too few jobs. Wait, let's go back.

The definition of a calorie is the energy needed to raise 1g of water 1'C.

WebElements: the periodic table on the WWW [http://www.webelements.com/]