it says "how many joules of energy would be required to heat 29.8 grams of carbon from 25.5 C to 79.3 C. (s= .71 Joules / grams times carbon) & the formula is Q= M x C x delta T .
9 answers
What's your problem? You have the formula and the numbers. Plug and chug.
How?
q = ?
m = mass = 29.8 grams
c = specific heat = .71 J/g*C
delta T = 79.3-25.5 = 53.8
q = 29.8 x 0.71 x 53.8 = ? joules.
It would be helpful if you told me why you couldn't do that, please.
m = mass = 29.8 grams
c = specific heat = .71 J/g*C
delta T = 79.3-25.5 = 53.8
q = 29.8 x 0.71 x 53.8 = ? joules.
It would be helpful if you told me why you couldn't do that, please.
So the answer would be 1233.5 joules?
Not according to my calculator. Did you punch in the right numbers?
Yeah. What did you get?
1138.3004?
You couldn't have.
29.8 x 0.71 x 53.8 = 1138.3 J. If you typed the problem in right and that's 0.71 the answer should be rounded to 2 s.f. which is 1.1E3 J. If that was really 0.710 then the answer should be rounded to 1.14E3 J.
It still would be helpful to know the real problem here. I hope this isn't a case of just not doing the work and getting someone on line to do it for you.
29.8 x 0.71 x 53.8 = 1138.3 J. If you typed the problem in right and that's 0.71 the answer should be rounded to 2 s.f. which is 1.1E3 J. If that was really 0.710 then the answer should be rounded to 1.14E3 J.
It still would be helpful to know the real problem here. I hope this isn't a case of just not doing the work and getting someone on line to do it for you.
1138.3004 is right. Now round it to either two or three s.f. depending upon how many you have in that 0.710 number.
1 answer