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
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
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?
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.
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.
To calculate the amount of heat required to raise the temperature of a substance, you need to use the formula:
Q = m * c * ΔT
where:
Q represents the amount of heat,
m is the mass of the substance,
c is the specific heat capacity of the substance,
ΔT is the change in temperature.
In this case, we have:
m = 25 g (mass of water)
c = 4.184 J/(g·°C) (specific heat capacity of water)
ΔT = (10.°C - 5°C) = 5°C (change in temperature)
Plugging in these values into the formula, we get:
Q = 25 g * 4.184 J/(g·°C) * 5°C
Q = 524 J
Therefore, the amount of heat required to raise the temperature of the water from 5°C to 10°C is 524 J. You rounded it to 520 J, which is incorrect as it does not account for the correct number of significant figures.
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.