Specific heat capacity
Investigate to determine the specific heat capacity of a metal cube provided
Using the normal equipment for Thermal Physics investigations, you are required to design and carry out an investigation to determine the specific heat capacity of a metal cube provide by teacher. Furthermore, you will need to compare your obtained value with published values, and confirm your identification of the metal by choosing a second physical property of the metal, and designing and carrying out an investigation to justify your findings.
Water= initial temperature: 24 , volume: 100ml
Final temperature: 28
Metal Block= initial temperature: 100, mass:10g
Final temperature : 28
Mass of calorimeter= 32g , Temperature same as water
Cw = 4200J×ng-1°C-1 Cal = 900kg-1°C-1
900×32×(28－24)＋4200×100×(28－24)=Cm×10×(100－28) Cm ＝ 115200 ÷70÷75
Cm ＝ 2194.2 J/kg
The numbers on the calorimeter is not nicety enough which leads the data not accurate. These matter might be get influenced by air temperature.
As you can see in the raw data, heat from the 100° metal cube has been transferred to the water by raising it initial temperature by 4°
Latent heat of fusion of ice
To determine the specific latent heat of ice by the method of mixture.
1. Weight the calorimeter together with stirrer, m0 by a beam balance 2. Heat some water in a beaker to about 40°C
3. Half fill the calorimeter with this water and weigh again m1 4. Record the room temperature T0
5. Allow the water in the calorimeter to cool, Meanwhile, dry the cubes of ice between dry towel 6. Take note of the temperature t1 of the water when it has dropped to about 10°C above the room temperature 7. Transfer the cubes of ice into the calorimeter as quickly as possible. Stir the micture gently and continuously until all the ics has melted. Record the temperature t2
8. Weight the calorimeter with its contents, m2
Water= volume: 100ml , at room temperature: 27
Temperature in the calorimeter : 39
Final Temperature(mixed with ice) : 15
Ice: temperature: 0 , volume: 11ml
Specific heat capacity of calorimeter = 900J/kgK
Specific capacity of water = 4200 J/kgK
Q absorbed = Q release
Qw + Qc = Qm
CwMw(Tfw-Tiw) + CcMc(Tfw-Tiw)
Q1 = Qw + Qe
=CwMw (Ti-Tf) + CcMc(Ti-Tf)
When we took the ice out I didn’t put the ice as soon as possible before it began to melting. Also, when we was pouring liquid from the beaker to the calorimeter, or the calorimeter to the beaker there might had some water we lost.
Estimating the size of a molecule with oil film method
-talcum powder baby powder
-large piece of glassboard
-A piece of coordinate paper
Through this experiment, let us learn a way of estimating the size of a molecule. The molecular formula of oleic acid is C17－H33COOH, and its molecule is made up of two parts: one is C17H33－, the other－COOH.－COOH has a strong affinity with water. When a drop of oleic acid id diluted by ethyl alcohol and dripped onto the surface of water, the acid drop spreads over the surface rapidly. The alcohol dissolves in the water and soon evaporates, and then a layer of pure oleic acid film forms. The part of C17H33 of the molecule sticks up from the water surface, but －COOH remains in water. The oleic and acid molecules erecting on the water surface form...
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