Chemical Energy Worksheet #1
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25.0 grams of a popular breakfast cereal contain 460 joules. How many joules are
released per gram? (This is called F.V. or Food Value which is found on food labels, read the
STSE unit on web site (not optional).
460 joules/25.0 grams = 18.4 joules/gram
A 55.6 gram sample of water is cooled from 50.0oC to 35.5oC.
a)
How many joules are released?
q = mct = 55.6 grams x 4.184 J/goC x (35.5oC - 50.0oC) = 3373 J
b)
How many kilojoules are released ?
3373 J x 1 kJ/1000 J = 3.3699 kJ
c)
How many calories would be released if 1 calorie = 4.184 joules?
3373 J x 1 calorie/4.184 J = 806.2 cal
A 22.2 gram sample of silver absorbs 65.6 J of heat energy when its temperature
is raised from 31.0oC to 43.5oC. Find the specific heat of silver. q = mct
65.6 J = 22.2 grams x ( c ) x (43.5oC - 31.0oC)
c = 0.236 J/g oC
How much heat is lost when 238.3 grams of H2O(g) at 100oC are converted to
liquid at 100oC? 40.8 KJ/mol is lost when stem condenses.
q = moles x molar heat of vaporization = 238.3 g x 1 mol/18.02 gr x 40.7 kJ/ mol
= 13.2 mol X 40.7 KJ/mol
q = 540 kJ
How much heat is lost when 12.4 grams of H2O(g) ar 110oC are converted to
liquid at 50.0oC? ( The specific heat of steam is 2.01 J/goC)
q = mct = 12.4 gr x 2.01 J/gr oC x ( 100oC - 110oC) = 0.249 KJ
q = moles x molar heat of vap = 12.4 gr x 1 mol/18 gr x 40.7 kJ/mol = 28.01kJ
q = mct = 12.4 x 4.18 J/gr oC x (50oC - 100oC) = 2.59 KJ
qt = 0.251 kJ + 28.01 kJ + 2.59 kJ = 30.9 kJ
A 5.0 kilogram block of ice at -10.0oC is placed in a container of warm water.
The block of ice is warmed to 0.0oC, and 4.0 kilograms of ice remains
unmelted. At this point, how many joules were transferred from the warm water
completely? (Molar heat to melt ice is 6.02 KJ/mol). (c for ice and steam same.)
q = mct = 5000gr x 2.03 J/gr oC x (-10oC - 0oC) = 101.5 KJ
q = moles x molar heat of fusion = 1000 g x 1 mol/ 18 gr x 6.02 kJ/mol
= 333.9 kJ
qt = 101.5 kJ + 333.9 kJ = 435 kJ
How much heat energy must be removed from a 45.0 grams sample of
naphthalene, C10H8, at its freezing point to bring about solidification? The
heat of fusion for naphthalene is 191.2 kJ/mol.
q = moles x molar heat of fusion = 45.0 gr x 1 mol / 128.18 g/mol x 191.2 kJ/mol
= 67.1 kJ
140.0 grams of H2O at 25oC is mixed with 100.0 grams of metal at 100.0oC. The
final temperature of the mixture is 29.6oC. What is the specific heat of the metal?
mwater x cwater x temp changewater = mmetal x cmetal x temp changemetal
140.0 gr x 4.18 J/groC x (29.6oC - 25oC) = 100.0 gr x c x (29.6oC - 100oC)
c = 0.383 J/g oC
If 596 J of heat are added to 29.6 grams of water at 22.9oC in a Styrofoam cup
calorimeter, what will be the final temperature of the water?
q = mct therefore
596 J = 29.6 gr x 4.18 j/gr oC x (Tf - 22.9oC)
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4.81 C = Tf - 22.9 so answer is Tf = 27.7oC