Psalm H.
asked • 10/26/16CHEM CONFUSION! HELP WITH CALORIMETRY
So the question is....
How many Joules of energy will be needed to heat 128.0g of water 0.00 degrees Celsius to 283.08 degrees K.
I keep getting 5320 J while my friends say that it is actually 5398 J! Which one is correct if any and why?
More
1 Expert Answer
Arturo O. answered • 10/27/16
Tutor
5.0
(66)
Experienced Physics Teacher for Physics Tutoring
Q = mcΔT
m = 128.0 g
c = 4.186 J/(g°C) [from tables]
T1 = 0.00°C
T2 = 283.08°K = (283.08 - 273.15)°C = 9.93°C
ΔT = T2 - T1 = (9.93 - 0)°C
Q = (128.0 g) [4.186 J/(g°C)] (9.93°C) = 5320.57 J
My result is close to your result.
Still looking for help? Get the right answer, fast.
Ask a question for free
Get a free answer to a quick problem.
Most questions answered within 4 hours.
OR
Find an Online Tutor Now
Choose an expert and meet online. No packages or subscriptions, pay only for the time you need.
Ismael A.
10/26/16