Question

Need help and an explanation on how to do this, please.
When 0.7521g of benzoic acid was burned in a calorimeter containing 1,000g of water, a temperature rise of 3.60 degrees C was observed. What is the heat capacity of the bomb calorimeter, excluding the water? The heat of combustion of benzoic acid is -26.42 kJ/g.

Answer 1

Is there an equation for you to plug these numbers into?

Answer 2

\[Q=(m)(c)(\Delta)(T)\]

Answer 3

I'm not too sure which one to use.

Answer 4

How come we use that equation even though there isn't really a change in temperature?

Answer 5

It says the temperature increased 3.6 degrees C so that is the temperature change

Answer 6

Ah, I see. But we don't have to calculate a temperature change in this case, right?

Answer 7

Nope! its already done for you! :)

Answer 8

The heat capacity would be "c" in the equation?

Answer 9

yes!

Answer 10

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?

Answer 11

If i'm correct, the water is just something thrown in there to mess you up, but Q is the -26.42

Answer 12

So this would the equation

-26.42= (0.7521)(c)(3.6)

Answer 13

C is the heat capacity

Answer 14

would the answer be -9.76J?

Answer 15

Thats what I got!

Answer 16

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 :(

Answer 17

would you like the link?

Answer 18

Sure!

Answer 19

https://answers.yahoo.com/question/index?qid=20120424181717AAxs958

Answer 20

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.

Answer 21

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!

Answer 22

i found 1 more explanation actually!

Answer 23

this person did it with less math

Answer 24

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

Answer 25

That looks less confusing but thats above what I know

Answer 26

Alright, great! Thank you.

Answer 27

No problem!