State of Matter is KILLA: HELP!!?
2) How many joules of energy are necessary to heat a sample of water with a mass of 46.0 grams from 0.0 Celsuis to 100.0 Celsius. (Use 4.184 J/g Celsius for the specific heat of water)
3) A sample of H2O with a mass of 46.0 grams has a temperature of 100 Celsius. How many joules of energy are necessary to boil the water? (Use 2.0934 J/g of the heat of vaporization of water.)
If anyone understands these three problems, you are a life saver because if I dont get these three questions right, Im going to fail and I wont be a senior next year. PLEASE ANYONE!!!
Answer:
For the first two questions, all you need to know is that
Temperature rise = energy supplied / (mass of object x specific heat of material)
or T = Q / ms
Whether it says "energy" or "heat" makes no difference here.
For your questions you need to rearrange the equation as
Q = msT
(1) T = 1000 – 20 = 980; m = 5.58; s = 450,000
Q = 980 x 5.58 x 450,000 = 246,078,000 J or 246.078 MJ.
(2) T = 100; m = 46; s = 4.184
Q = 100 x 46 x 4.184 = 19,246.4 J or 19.2464 kJ.
For the last question, you need to know that it takes energy to turn water into steam. This is called the (latent) heat of vaporization.
(3) Mass of water = 46 g. Energy to boil 1 g of water is given as 2.0934 J.
So energy to boil 46 g is 46 x 2.0934 = 96.2964 J.
Good luck!
for 1 and 2 use heat=mass x specific heat x change in temperature
3 use heat= mass x heat of vaporisation
Just use -:
Q = m c deltaT
1) Q = m c deltaT
Q = 5.58 x 450,000 x (1000 - 20)
Q = 2.0088 x 10^8 Joules
2) Q = m c deltaT
Q = 46 x 4.184 x (100 - 0)
Q = 19246.4 Joules
3)Q = m c deltaT
Q = 46 x 2.0934 x 1 (i.e. no change in temp)
Q = 96.2964 Joules
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