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Water and Ice- To determine the specific heat capacity of water (a) and the specific latent heat of fusion of water (b).

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Physics Full Lab Report (1) Title: (B3) Water and Ice- To determine the specific heat capacity of water (a) and the specific latent heat of fusion of water (b). Date: 25 October 2004 Aim: To determine the specific heat capacity of water through electrical method and to determine the specific latent heat of fusion of water using the method of mixture. In the end, the seriousness of heat loss from the heating curve of water by a heater is to be judged. Experiment (a): To determine the specific heat capacity of water Principle: Specific heat capacity of a substance is defined as E= mc?T, where the energy E depends on the mass m of the object and the temperature change?T. For determination, a calorimeter, which consists of an aluminum calorimeter placed inside an insulating jacket made polystyrene, is used. A certain amount of water is added inside the calorimeter and heated with a heating coil connected to a circuit with power supply. ...read more.


Final temperature of water 47.1? Temperature change 5.1? Total time elapsed 2080 s 1. 2. By E=VIt, the total energy evolved during heating of heating coil is: E=(2.6)(2)(2080) E=10,816 J 3. Then by subsidizing the required data into (ii), we have: 10816=(100/1000)c(5.1)+(30.9/1000)(900)(5.1) c=20,930 J kg-1 K-1 Discussion: 1. Compared to the standard value of specific heat capacity of water, 4200 J kg-1 K-1, the percentage error is (20930-4200)/4200x100%�398% 2. Such enormous error cannot be accepted. Some unexplainable mistakes or faultiness must have occurred during procedures. Time, however, does not allow us to re-do the experiment. Thus, the experimental data remains valid. Experiment (b): To determine the specific latent heat of fusion of water Principle: An object absorbs latent heat of fusion while it melts. Specific latent heat of fusion can be found by: E=mlf, where energy E depends on the mass m of the object. In this experiment, some hot water heated ?10? the room temperature in the calorimeter is to be cooled down by adding ice cubes into the until the temperature falls to 10? ...read more.


and (iv), we have E1=(100/1000)(4200)(39.5-15)+(30.9/1000)(900)(39.5-15) E1=10,971.345 J E2=E1 E1=(59/1000)l+(59/1000)(4200)(0-15) l f=248,955 J kg-1 Discussion: 1. As the ice cubes were inserted into the hot water, its temperature falls gradually. 2. The magnitude of temperature fall of the hot water for each ice cube is proportional to the size of the ice cube to be added. 3. The resulting value of specific latent heat of fusion of water, when compared with its standard value of 334,000 J kg-1, is underestimated by: (334000-248955)/334000 x 100%�25.5% Sources of error: 1. Heat is lost to the surroundings by convection, conduction and radiation. 2. Heat absorbed by the polystyrene foam in the jacket is not measured. 3. Thermometer is inaccurate in scale, thus errors may occur in measuring temperature with ordinary thermometer. 4. Personal errors may occur while taking thermometer readings and weighing objects as well. Conclusion: According to the comparison between the standard values and the experimental results, it is obvious that the methods used in their corresponding experiments cannot give accurate results. The main reason is believed to be due to the accounting of serious unmeasured heat loss and errors of system, personal, etc. 1 ...read more.

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