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Is there an equation for you to plug these numbers into?
I'm not too sure which one to use.
How come we use that equation even though there isn't really a change in temperature?
It says the temperature increased 3.6 degrees C so that is the temperature change
Ah, I see. But we don't have to calculate a temperature change in this case, right?
Nope! its already done for you! :)
The heat capacity would be "c" in the equation?
so what i'm not really understanding is: they've given the mass (0.7521g of benzoic acid) and they've given the temperature (3.60 degrees C).. where/how do we use the 1,000g of water and the heat of combustion of benzoic acid in the question?
If i'm correct, the water is just something thrown in there to mess you up, but Q is the -26.42
So this would the equation -26.42= (0.7521)(c)(3.6)
C is the heat capacity
would the answer be -9.76J?
Thats what I got!
I did find an explanation to this problem on yahoo answers (and i'm not doubting your way) but they got another answer and used steps i don't understand :(
would you like the link?
Woah. Thats a lot of work. I checked the guys bio and he is a chemistry professor so i would probably go with his answer, but I assumed that it would be the explanation I gave you. I don't understand any of that, but then again he is a professor. You could write both solutions down and then ask your teacher which one is right. If your class has been doing Specific Heat and Heat Capacity equations then I would use mine, but if that is not the case then the other guy is probably correct with is crazy math.
ha! That's how i felt :( I'll definitely ask on both because not knowing the way he did on yahoo answers means we didn't go over it in class but yours made sense as well.. even though now i'm double confused lol but thank you for your help!
i found 1 more explanation actually!
this person did it with less math
Q(comb) = (0.7521g)((-26.42KJ/g) = -19.87KJ Q(H20) = (3.6C)(1000g)[4.184J/(g-C)] = 15.062KJ Q(cal) = 19.87KJ - 15.062KJ = 4.81KJ SpH = 4.81KJ/3.6C = 1.34 KJ/C
That looks less confusing but thats above what I know
Alright, great! Thank you.