Introduction
The aim of this investigation is to determine the specific heat capacity of H2O (water). This is accomplished by measuring to which temperature a certain amount of energy takes the water. The independent variable of this experiment is the amount of current put into the water. The dependent variable is the water’s temperature over time. The controlled variables are the initial temperature of the water and the amount of water.
Materials and Methods
The following materials were used:
- scale
- calorimeter
- stirrer
- thermometer
- stopwatch
- power supply
- multimeter
- 385 g H2O
The calorimeter was filled with 385 g of water, after which a measured current was applied to it the water. The temperature of the water was measured in increments of one minute, while the water was stirred with the stirrer.
The experiment was conducted only once, and thus the independent variable was not varied.
Results
| Table 1: Measured current |
| Voltage (V) |
15.81 V |
| Current (I) |
3 A |
| Power (P) |
47.43 W |
P (the power) was derived with the following formula:
P = V * I
| Table 2: Temperature over time |
| Time [m] |
Temperature [C] +/- 1 |
| 0 |
16 |
| 1 |
16 |
| 2 |
18 |
| 3 |
21 |
| 4 |
22 |
| 5 |
25 |
| 6 |
27 |
| 7 |
28 |
| 8 |
30 |
| 9 |
31 |
| 10 |
33 |
| 11 |
35 |
| 12 |
36 |
| 13 |
38 |
| 14 |
40 |
| 15 |
41 |
| 16 |
43 |
| 17 |
44 |
| 18 |
46 |
| 19 |
48 |
| 20 |
49 |
| 21 |
51 |
| 22 |
52 |
Since an accurate stopwatch was used, the error margin in the time-column is negligible. The thermometer used was difficult to measure with because of unclear display, which approximates to an error margin of 1 degree Celsius. There is a clear, linear trend visible in the data.
The specific heat capacity (c) was calculated using the following formula:
C = Q / (m*dT)
Where m is the mass of the water and dT is the change in temperature. Q is the heat added, which equals P times the time passed. Since the trend in the data is regular and lacks anomalous results, the total data can be used for this calculation. The calorimeter’s effect was neglected.
c = Q / ( m * dT ) = ( 47.43 * 60 * 22 ) / ( 0.350 * ( 52 - 16 ) ) = 4969 J(kgK)^-1
Conclusion
According to this experiment, the specific heat capacity of liquid water at 16 degrees Celsius is 4969 J(kgK)^-1. Comparing our result with literature values, Giancoli’s Physics (fifth edition) states that this value is 4186 J(kgK)^-1. Since our result was larger than Giancoli’s, we must have disregarded another loss of heat. This is because we assumed that the water absorbs all heat, which is not the case. Some heat is lost to the surroundings, e.g. the calorimeter and the air above it.
Note, also, that the current varied slightly during the investigation. It started as 3 A, but when the experiment had been completed, it had risen by approximately 0.1 A. From this, we can draw the conclusion that the resistance of water decreases by a small amount when it is heated between 16 and 52 degrees Celsius.
Evaluation
The method used is flawed on several points. The main weakness, though, is the fact that energy is lost to the environment. The calorimeter was a polystyrene cylinder with a mass of approximately 0.2 +/- 0.1 kg. Polystyrene has a specific heat capacity of 1.3 times 10^3 J(kgK)^-1 according to Wikipedia. The c*m factor of the formula for heat added (derived from the formula for specific heat capacity) is then for polystyrene 1.3 times 10^3 times 0.2 = 260 JK^-1, while the respective factor for water is 4.969 times 10^3 times 0.350 = 1739 JK^-1. Thus, the polystyrene calorimeter stands for a significant part of the heat lost. Heat was also lost to the heating apparatus as well as to the stirring stick and the thermometer, although these objects had low masses. By using an even more0 insulating calorimeter, with a lower specific heat capacity, we would decrease the heat lost to the environment.
The primitive method for stirring is another reason for heat-loss. A stick with a ring in attached to the end was used to manually stir the water, which led to exhausted arms and an inability to maintain stirring for a prolonged period of time. This could have been solved by using an automatic stirrer.
Introduction
The aim with this investigation is to verify Newton’s second law of motion by exerting a force on a body and measuring its acceleration. The independent variables in this experiment are the force exerted and the mass of the body. Acceleration is the dependent variable.
Materials and Methods
An object was put on an apparatus which creates a near-frictionless surface. A force was applied to the object. We measured the object’s momentaneous velocity on two points along the surface, as well as the time elapsed during traveling between those two points.
The independent variables, the mass of the weight and the mass of the glider’, are varied throughout the experiment. The experiment was conducted four times with every set of independent variables.
Results
The distance measured on t1 and t2 was 1 dm.
| Table 1: Glider results |
| glider mass [kg] |
weight mass [kg] |
t1 [s] |
t2 [s] |
t1 to t2 [s] |
| 0.360 |
0.010 |
0.35885 |
0.17230 |
1.27652 |
| 0.360 |
0.010 |
0.37776 |
0.17312 |
1.31245 |
| 0.360 |
0.010 |
0.35900 |
0.17135 |
1.27624 |
| 0.360 |
0.010 |
0.42935 |
0.17702 |
1.39867 |
| 0.522 |
0.010 |
0.43511 |
0.20445 |
1.53560 |
| 0.522 |
0.010 |
0.43380 |
0.20491 |
1.53196 |
| 0.522 |
0.010 |
0.41020 |
0.20373 |
1.49012 |
| 0.522 |
0.010 |
0.43690 |
0.20686 |
1.155194 |
| 0.522 |
0.030 |
0.25305 |
0.12068 |
0.903039 |
| 0.522 |
0.030 |
0.25579 |
0.12127 |
0.907210 |
| 0.522 |
0.030 |
0.32114 |
0.12627 |
1.02483 |
| 0.522 |
0.030 |
0.24598 |
0.12077 |
0.885258 |
| 0.763 |
0.030 |
0.40276 |
0.15211 |
1.25234 |
| 0.763 |
0.030 |
0.29374 |
0.14348 |
1.05475 |
| 0.763 |
0.030 |
0.30457 |
0.14473 |
1.07882 |
| 0.763 |
0.030 |
0.31121 |
0.14616 |
1.09995 |
| 0.360 |
0.080 |
0.13825 |
0.06657 |
0.49296 |
| 0.360 |
0.080 |
0.13440 |
0.06683 |
0.49534 |
| 0.360 |
0.080 |
0.12847 |
0.06568 |
0.47251 |
| 0.360 |
0.080 |
0.13646 |
0.06675 |
0.49048 |
| 0.749 |
0.080 |
0.19572 |
0.09245 |
0.69013 |
| 0.749 |
0.080 |
0.18879 |
0.09160 |
0.67628 |
| 0.749 |
0.080 |
0.18645 |
0.09101 |
0.67149 |
| 0.749 |
0.080 |
0.19026 |
0.09175 |
0.67893 |
No anomalous results were recorded.
The acceleration was calculated twice for every set of data: First the estimated acceleration, calculated with Newton’s second law of motion, and then the real acceleration, calculated with the data from the photocells. This is an example calculation for the first set of data:
Acceleration Calculated with Newton’s Second Law of Motion
Fnet = mweight X G = 0.01kg X 9.82ms-2 = 0.0982N
Fnet = a X m => a = Fnet / m
=> a = 0.0982N / 0.370kg = 0.265ms-2
Acceleration Measured
a = (xchange) / t = ( d/t1 - d/t2 ) / t
=> a = ( 0.581 - 0.279 )ms-1 / 1.277s
=> a = 0.236ms-2
The distance in the first row is the distance that the photocells measured time of. It was equal to 0.1 m.
Calculated Values
| Table 2: Calculated and measured results |
| glider mass [kg] |
weight mass [kg] |
t1 [s] |
t2 [s] |
t1 to t2 [s] |
calculated acceleration [ms-2] |
measured acceleration [ms-2] |
| 0.360 |
0.010 |
0.35885 |
0.17230 |
1.27652 |
0.265 |
0.236 |
| 0.360 |
0.010 |
0.37776 |
0.17312 |
1.31245 |
0.265 |
0.239 |
| 0.360 |
0.010 |
0.35900 |
0.17135 |
1.27624 |
0.265 |
0.240 |
| 0.360 |
0.010 |
0.42935 |
0.17702 |
1.39867 |
0.265 |
0.187 |
| 0.522 |
0.010 |
0.43511 |
0.20445 |
1.53560 |
0.185 |
0.170 |
| 0.522 |
0.010 |
0.43380 |
0.20491 |
1.53196 |
0.185 |
0.168 |
| 0.522 |
0.010 |
0.41020 |
0.20373 |
1.49012 |
0.185 |
0.165 |
| 0.522 |
0.010 |
0.43690 |
0.20686 |
1.155194 |
0.185 |
0.220 |
| 0.522 |
0.030 |
0.25305 |
0.12068 |
0.903039 |
0.534 |
0.478 |
| 0.522 |
0.030 |
0.25579 |
0.12127 |
0.907210 |
0.534 |
0.481 |
| 0.522 |
0.030 |
0.32114 |
0.12627 |
1.02483 |
0.534 |
0.437 |
| 0.522 |
0.030 |
0.24598 |
0.12077 |
0.885258 |
0.534 |
0.475 |
| 0.763 |
0.030 |
0.40276 |
0.15211 |
1.25234 |
0.372 |
0.327 |
| 0.763 |
0.030 |
0.29374 |
0.14348 |
1.05475 |
0.372 |
0.340 |
| 0.763 |
0.030 |
0.30457 |
0.14473 |
1.07882 |
0.372 |
0.335 |
| 0.763 |
0.030 |
0.31121 |
0.14616 |
1.09995 |
0.372 |
0.330 |
| 0.360 |
0.080 |
0.13825 |
0.06657 |
0.49296 |
1.785 |
1.557 |
| 0.360 |
0.080 |
0.13440 |
0.06683 |
0.49534 |
1.785 |
1.508 |
| 0.360 |
0.080 |
0.12847 |
0.06568 |
0.47251 |
1.785 |
1.552 |
| 0.360 |
0.080 |
0.13646 |
0.06675 |
0.49048 |
1.785 |
1.592 |
| 0.749 |
0.080 |
0.19572 |
0.09245 |
0.69013 |
0.948 |
0.836 |
| 0.749 |
0.080 |
0.18879 |
0.09160 |
0.67628 |
0.948 |
0.825 |
| 0.749 |
0.080 |
0.18645 |
0.09101 |
0.67149 |
0.948 |
0.836 |
| 0.749 |
0.080 |
0.19026 |
0.09175 |
0.67893 |
0.948 |
0.826 |
Conclusion
The calculated values are clearly related to the measured values: they are around 10% higher in the majority of the cases. This means that Newton’s second law of motion probably is correct, but there might be a systematic error. That error is omission of friction. The so-called frictionless surface was not entirely frictionless, which also explains why the error margin is greater if the glider passed the photocells late in its course rather than early.
Evaluation
This method is fairly viable, but it has one weaknesses: Friction is not accounted for, neither to the surface nor to the air, which affects both the weight and the glider. This could be fixed by performing the investigation on an entirely frictionless surface in vacuum. However, this is difficult to achieve.
The effect of this problem can also be decreased by using a heavier weight and a lighter glider.
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