 ```Energy Worksheet
(spend up to 30 minutes on this then stop)
1.
25.0 grams of a popular breakfast cereal contain 460 joules. How many joules are
released per gram?
460 joules/25.o grams = 18.4 joules/gram
2.
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 4.18 J/goC x (35.5oC - 50.0oC) = 3369.9 J
b)
How many kilojoules are released?
3369.9 J x 1 kJ/1000 J = 3.3699 kJ
c)
How many calorie would be released if 1 calorie = 4.18 joules?
3369.9 J x 1 calorie/4.18 J = 806.2 cal
3.
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 = .236 J/gr oC
4.
How much heat is lost when 238.3 grams of H2O(g) at 100oC are converted to
liquid at 100oC?
q = moles x molar heat of vaporization = 238.3 gr x 1 mol/18 gr x 40.7 kJ/ mol
q = 538.8 kJ
5.
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 2.02 J/goC)
q = mct = 12.4 gr x 1.86J/gr oC x ( 100oC - 110oC) = -230.6 J
q = moles x molar heat of vap = 12.4 gr x 1 mol/18 gr x 40.7 kJ/mol = 28.04kJ
q = mct = 12.4 x 4.18 J/gr oC x (50oC - 100oC) = 2591.6 J
qt = .231 kJ + 28.04 kJ + 2.59 kJ = 30.9 kJ
6.
A 5.0 kilogram block of ice at -10.0oC is placed in a container of warm water.
The entire 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?
q = mct = 5000gr x 2.09 J/gr oC x (-10oC - 0oC) = - 10450 J
q = moles x molar heat of fusion = 1000 gr x 1 mol/ 18 gr x 6.01 kJ/mol
= 333.9 kJ
qt = 10.45 kJ + 333.9 kJ = 344.3 kJ
7.
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 gr x 191.2 kJ/mol
= 67.2 kJ
8.
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 = .382 J/gr oC
9.
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
596 J = 29.6 gr x 4.18 j/gr oC x (Tf - 22.9oC)
4.81oC = Tf - 22.9
Tf = 27.7oC
```