Chemistry
posted by Jane .
Calculate how much heat 32.0 g of water absorbs when it is heated from 26.0°C to 73.5°C.
In joules, and then we have to calculate calories too, how is that done? I was absent. Thanks so much!

q = mass x specific heat x (TfinalTinitial).
For joules, use 4.184 J/gram*C for specific heat. For calories, use 1.00 cal/gram*C 
What's the constant for water though? Every time I look it up it's a different number

Oh nevermind you put it there. THANK YOU SO MUCH! You're a huge help!!!

I gave you the two constants. If you want q in joules, use specific heat = 4.184 J/g*C and if you want q in calories, use specific heat = 1.00 calorie/g*C.
You can look up specific heat in J/g, J/mol, cal/g, cal/mol, kg/mol, kg/g, etc. I gave you the two constants you need. 
Thank you so much!!! You're really a lifesaver!

Is this is food calories?

no. Years ago the food industry got it all wrong. They talked about Big calories (identified by a CAPITAL letter C) and a small calorie (identified by a lower case c). But they made no real distinction and soon everyone (perhaps I should say the nonprofessionals) just called them "calories". A food calorie (the Big Calorie) actually is a kilocalorie. So when I drink a glass of nofat milk, the label on the container says it is 80 calories. That is 80 kilocalories. So if we are to eat about 2000 calories a day to maintain our weight, that actually is 2000 kcal or 2,000,000 calories.

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