CHM130
Lab 4
Calorimetry
Name:
Data Table: (12 points)
ALUMINUM METAL
Pre-weighed Aluminum metal sample mass (mmetal)
20.09 g
Temperature of boiling water and metal sample in the pot (Ti(metal))
dsdfa(Ti
99°C
Temperature of cool water in the calorimeter prior to adding hot metal sample (Ti(water))
24°C
Maximum Temperature of water/metal in calorimeter after mixing (Tf)
28°C
LEAD METAL
Pre-weighed Lead metal sample mass (mmetal)
20.03g
Temperature of boiling water and metal sample in the pot (Ti(metal))
103°C
Temperature of cool water in the calorimeter prior to adding hot metal sample (Ti(water))
25°C
Maximum Temperature of water/metal in calorimeter after mixing (Tf)
26°C
IRON METAL
Pre-weighed Iron metal sample
mass
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You have the mass of water from calculation #9, the specific heat of water is 4.184 J/g(oC), and the temperature change of water from calculation #7. (10 points)
Mass of water: 74.8g
Specific heat of water is 4.184 J/g(oC)
Change in temp. of water: 1°C
Q = 74.8 * 1 * 4.184 = 312.96 J/g(oC)
11. Use the equation: q = m(SH)ΔT to solve for the specific heat of the metal.
For q, you found how much heat was gained by the water so you know that same amount of heat was lost by the metal. Therefore, qmetal = -qwater. The mass of the sample was recorded from the baggie. The temperature change of the metal can be found in calculation #8. (10 points)
Mass of Sample: 20.03 g
Change in temperature of metal: 77°C
Q = 312.96 J/g(oC)
SH = q/(m* ΔT):
SH = (312.96) / (20.03 * 77) = 312.96 / 1542.31 = 0.2029°C
12. Determine the percent error using the equation and knowing that the actual specific heat of lead is 0.130 J/g(oC): (10 points)
Percent Error = actual-experimental x 100
Actual
Percent Error: (0.130 – 0.203)/ (0.130) * 100 = -56.15% error
Iron Sample:
13. Calculate the change in temperature for the water caused by the addition of the aluminum by subtracting the initial temperature of the water from the final temperature of the water. ΔT = Tf - Ti (5 points)
ΔT = 26°C – 25°C = 1°C
14. Calculate the change in the metal’s temperature by subtracting the initial temperature of the iron from the final
The dependent variable in the experiment was the temperature and energy absorbed by the water.
The additional ice that melted was added to the initial mass of water to obtain the final weight of water in the solution to calculate molality:
3. The volume of a fixed mass of a liquid sample increases as the temperature rises from 20 to
In this experiment, we investigate the change in temperature caused by adding a chemical substance into the water and dissolving it. The results recorded in the table below show that our hypothesis is correct.
The average calorimeter value as calculated was -842.325 j/0c. Negative values of temperature change signify heat loss. The initial and final temperatures are 25.60C and 45.50C respectively. This suggests that, heat was transferred from the hot to the cold water as well as from the hot metal to the calorimeter. Thus, the outcomes of the experiment outcome were consistent with what was hypothesized.
The hypothesis in this experiment that will be tested is “ If there is a higher temperature inside the jar,then bigger crystals will form because the heat speeds up reactions.’To figure out if this is correct, the scientist must get the temperature of each of the conditions they will be testing in. This composes
Furthermore, we can calculate the molar mass of the metal by using this equation: cmetal * MMmetal ≅ 25 J/mol. °c In this report I am going to use the SI unit of the energy, which is joule. 1 joule is equal to 1 kg.m2/s2.
b) An empty beaker was weighted. Then, water was filled in the beaker. The temperature was recorded at uniform intervals.
Mass of water x 4.2 (water’s specific heat capacity) x temperature change = energy transferred from the fuel to the water
Introduction: The theory behind this experiment is the heat of a reaction (∆E) plus the work (W) done by a reaction is equal to
c = the specific heat capacity of water (this is 4181 J / Kg oC)
2. The end of step 4 during the experiment creates the tripartite thermal structure within the lake substitute. This is composed of three parts called the epilimnion, the thermocline and the hypolimnion. The epilimnion represents the surface of the lake which experiences the most heat during the summer and has the highest temperature compared to the other parts of the thermal structure. The thermocline is the section of water that has the large exponential decrease in temperature. This is shown in figure 1 at the end of the heating phase, the temperature decreases by a large amount as the depth increases from 1cm to 9cm. Finally there is the hypolimnion which is shown in figure 1 in all phases as the cold part of the water column. This is the most unaffected part, as the solar radiation does not effectively penetrate that deep into the water hence the temperature stays almost constant. The
(a) The quality of the mixture is 0.6 which means 60% of the mass of mixture is in vapour form while 40% of the ass is still in liquid phase. This means this is a saturated liquid-vapour mixture. We can now refer to the saturated water and steam table to find the temperature using the pressure given to us.
The objective of the experiment was to measure and determine the coefficient of thermal conductivity for copper and stainless steel.
Specific heat of a metal, cs = mH2O cH2O (Tf – Ti) / ms (100oc – Tf)