IV. Results:
The following table can be used to determine the density of a liquid.
Mass of clean dry
graduated
cylinder
(grams)
Volume of liquid
(ml)
Mass of cylinder
and liquid
(grams)
New Volume of
liquid
ml
New mass of
cylinder and
liquid
(grams
Liquid A
23.30g
Liquid B
23.50g
Liquid C
23.16g
6.2 ml
6.1ml
6.5ml
28.03g
31.33g
29.71g
8.2ml
9.0ml
8.8ml
29.86g
34.30g
31.89g
D=m = (28.03g-23.30g) = .7629 g/ml = .76 g/ml
V
(6.2 ml)
Average = .78 g/ml
2-propanol
Average = 1.2 g/ml
1,2,3 propanetriol
D=m = (29.86g-23.30g)= .8 g/ml = .80 g/ml
V
(8.2 ml)
D=m = (31.33g-23.50g) = 1.2836 g/ml = 1.3 g/ml
V
(6.1 ml)
D=m = (34.30g-23.50g)= 1.2 g/ml
V
(9.0 ml)
D=m = (29.71g-23.16g) = 1.007 g/mL = 1.0 g/mL
V
(6.5 ml)
D=m = (31.89 g-23.16g)= .992 g/ml = .99g/mL
V
(8.8 ml)
Average = 1.0g /mL water
Specific Heat Capacity Lab Calculations
IV. Results:
The following data can be used to determine the specific heat capacity of a piece of metal.
Metal = Iron
Mass of metal (grams)
Initial temperature of the
metal
(Temperature of the
boiling water)
Final Temperature of
metal ( Highest
temperature of the water
achieves after the hot
metal is placed in it. )
I.
Water
Mass of water in
the Styrofoam cup
Initial temperature
of the water in the
Styrofoam cup
52.67g
100 oC
31.2 oC
52.7 ml = 52.7 ml
23.1oC
31.2oC
Final temperature
of the water in the
Styrofoam cup
after the hot metal
is placed in it.
Conclusion:
1. a) The specific heat capacity of the water is 1.00 cal / g oC or 4.18 J / g oC
Water requires 1 cal to raise 1 gram of it 1oC.
b) The heat lost by the metal equals the heat gained by the water
Heat lost = mass of metal ( temperature change of metal) specific heat capacity of metal =
mass of water ( temperature change of water ) specific heat capacity of water = Heat gain
QL = mmetal ( tinitial of
metal
– tfinal of metal ) Cmetal = mwater ( tfinal of
water
– tinitial of water) Cwater = QG
mm ( tim – tfm ) Cm = mw ( tfw – tiw) Cw
Cm
Cm
=
mw ( tfw – tiw) Cw
mm ( tim – tfm )
= 52.7g (30.1oC-23.1oC) ( 1cal / g oC)
52.67g (100oC-30.1oC)
cm = .10 cal / g oC
IV. Results Table: The following data can be used to determine the latent heat of fusion of ice.
mass of Styrofoam cup = 10.26g
mass of cup and room temperature water 77.24g
initial temperature of the room temperature water
23.0oC
equilibrium temperature of the water after the ice melts 4.5oC
mass of calorimeter and water after the ice has melted. 91.86 g
V. Conclusions:
Q = m c (Tfinal-Tinitial)
c= 4.180 J / goC of water c= 1 cal / goC
1. Determine the heat lost by the room temperature water in calories.
Q = m c (Tfinal-Tinitial) = ( 77.24g – 10.26g)( 4.5oC-23.0oC) (1 cal / goC)= 791 cal = 7.9x102 cal
2. After the ice melts it is liquid water with a temperature of 0 oC and a specific heat of 1000cal/kgoC. or
4180 J / kgoC.
This water gains energy until it reaches the equilibrium temperature.
Determine the amount of heat gained by the ice ( water after it melts) in calories.
Q = m c (Tfinal-Tinitial) = ( 85.86g – 77.24g)( 4.5oC-0.0oC) (1 cal / goC)= 39 cal
3. Using the principle of conservation of energy determine the amount of heat absorbed by the
melting ice in cal and joules
7.9x102 cal – 39 cal = 751 cal = 7.5x102 cal
4. Determine the Latent Heat of Fusion of Ice in cal per g and J per kg
Hf = 7.5 x102 cal = 87 cal/g
87cal( 4.18 J) ( 1000g ) = 3.6 x106 J
( 85.86g – 77.24g)
g ( 1cal ) ( 1 kg )
IV Results: The following data can be used to determine the Latent Heat of Vaporization of water
mass of cup=10.26g
mass of cup and water=111.30g
initial temperature of water(from graph)=32.6 oC
initial time=101s
final temperature of water(from graph)=64.8oC
final time=298.5s
time started boiling(from graph)=810s
time stopped boiling=999s
new mass of cup and water after boiling=98.60g
V. Conclusion:
a) What is the rate of heat transfer from the hot plate to the water?
Rate = mass of water * factor to change g to Kg * temperature change * specific heat capacity
time needed cause temperature change
Rate = (111.30g -10.26g)*(1kg/1000g)*( 64.8oC -32.6 oC)*(4180 J / kg oC) = 68.9 J / s
(298.5s –101s)
b) How many joules were needed to boil the water?
Heat to vaporize = time boiling * rate of heat transfer
Heat = (999s -810s)*(68.9 J/s) = 13,022 J or 1.30x104J
c) What is the heat of vaporization of the water in J/kg?
Heat of Vaporization =
Heat / kg =
Heat needed to boil divided
by mass vaporized converted from grams to kilograms
1.30x104 J
= 1,023,622 J / kg = 1.02 x106J / Kg
[(111.30g –98.60g)*( 1 kg / 1000g)]
II.
Results: The following data can be used to determine the number of calories released
per gram of peanut.
Mass of peanut = .42 g
Mass of Al can = 22.16 g
Mass of Al can and water = 74.56 g
Initial temp of Al can and water 23.10C
Final temp of Al can and water 41.20C
cwater= 1.0 cal/g oC
cAl= .21 cal/g oC
1000 cal = 1 kcal
1000 J = 1 kJ
1 cal = 4.18 J
1000 g = 1kg
III.
Calculations / Conclusions
Heat released by nut
Q= (74.56g-22.16g) ( 1.0 cal/g oC) ( 41.20C_ 23.10C)= 948 cal
Q= (74.56g-22.16g) ( .21cal/g oC) ( 41.20C_ 23.10C)= 199 cal
Q= 1147 cal = 1.15x103 cal
2. Heat released in cal per gram of peanut
1.15x103 cal = 2738 cal = 2.74 x103 cal
.42 g
g
g
3. Heat released in kcal per gram
2.74 x103 cal ( 1 kcal ) = 2.74 kcal
g (1000cal)
g
4. Heat released in J per gram
2.74 x103 cal ( 4.18 J ) = 11,543 J =1.15x104 J
g (1 cal)
g
g
5. Heat released in KJ per gram
1.15x104
J
g
( 1 kJ) = 11.5 KJ
(1000J)
g
Big Alka-Seltzer
Results: The following can be used to determine the number of calories per gram
absorbed from the reaction of citric acid and sodium hydrogen carbonate
Mass of cup = 10.62g
Mass of cup + citric acid = 20.63g
Mass of cup , water and citric acid =70.54g
Initial temperature of mixture =23.1oC
Mass of sodium hydrogen carbonate=10.02g
Final temperature of mixture =5.2oC
Specific heat capacity of water = 1.0 cal / g oC
1 Cal = 1 kcal = 1000 cal
1 cal = 4.18 J
1 kJ = 1000 J
IV.
Calculations / Conclusions
1. The heat was absorbed by the reaction from the water,
The amount of energy absorbed by the reaction from the water substance is dependent on its
specific heat capacity, its mass, and its temperature change.
Heat absorbed = specific heat capacity of water * mass of water* temperature change
Q = [(1 cal / g oC)*( 70.54 g -20.63 g )*(23.1 oC -5.2oC)] = 893 cal = 8.9 x102 cal
2. Determine the number of calories per gram reacted.
8.9 x102 cal
= 44 cal
[(20.63g-10.62g)+10.02g]
g
3. Convert the number of calories per gram to Calories or kcal per gram.
44 cal ( 1 kcal ) = .044 kcal
g 1000 cal
g
4. Convert the number of cal per gram to J per gram.
44 cal ( 4.18 J ) = 184 J = 1.8 x102 J
(1 cal )
g
5. Convert the number of cal per gram to KJ per gram
1.8x102J ( 1 kJ ) = .18 KJ
(1000J)