Calculate the amount of heat required to raise the temperature of a 25 g sample of water from 5°C to 10.°C.
i got 523 then rounded it to two sig figs
520 but it said it was wrong
9 answers
Good. You rounded BUT 520 is to three s.f. I would round it to two but write it as 5.2 x 10^2 J. Are you sure it's two s.f.? 10-5 is to only one s.f. which would be 5 x 10^2 J.
thanks!
i have another question:
A 43 kg sample of water absorbs 343 kJ of heat. If the water was initially at 22.1°C, what is its final temperature?
I got 1.9 but i think i didn't do it correctly
A 43 kg sample of water absorbs 343 kJ of heat. If the water was initially at 22.1°C, what is its final temperature?
I got 1.9 but i think i didn't do it correctly
1.9 is the difference in Tfinal-Tinitial so Tfinal must be 24.0 (24.0-22.1 = 1.9).
343,000 = 43,000 x 4.184 x (Tf-22.1) = 0
Solve for Tf which I get as 24.0.
343,000 = 43,000 x 4.184 x (Tf-22.1) = 0
Solve for Tf which I get as 24.0.
thank you so much!!
oh, would it be 24 (two sig figs) cause in the problem 43 has the least number of sig figs(2) or do we go by this equation:
343,000 = 43,000 x 4.184 x (Tf-22.1) = 0
and use 3 sig figs?
343,000 = 43,000 x 4.184 x (Tf-22.1) = 0
and use 3 sig figs?
Hannah, Rebekah -- it's hard to remember which name to use when you use more than one, isn't it?
Please stick with just one name in this forum.
Please stick with just one name in this forum.
I think you are right. 43 has only two s.f. so the answer should be 24. Technically we don't know how many s.f. are in 43 kg. If they meant two it should have been written as 4.3 x 10^1; still, I would go with two and round the final answer to 24.
And Ms. Sue has a very good point. I saw the same question above under Hannah and thought I had a different student (until I came back here to copy the answer I had given to Rebekah).