Thermal Capacity, the Amount of Heat to Be Supplied to a Given Mass of a Material - Research Paper Example

Heat Name Student Number FY014 Date of Experiment Date of Submission Abstract This report describes an experiment to determine the heat capacity of a block and thermal conductivity of a rod. Observable parameters including temperature and mass were used to calculate the heat capacity of a block and thermal conductivity of a rod used. The experiment consisted of two parts. The first part describes an experimental method of determining specific heat capacity of a block used. The second part describes an experimental method of determining thermal conductivity of a rod used. In both cases, graphs were plotted to assist in discussions and analysis of the results obtained. The main aim of the experiment was for familiarization with using various laboratory apparatus, taking readings accurately using these apparatus and performing various computations on the values obtained. The report describes the setup of the experiment; the results recorded, discussed and analyzed. Moreover, calculated values were compared to the expected values. 1.0 Introduction Heat is defined as the transfer of energy across the boundary of a system due to a temperature difference between the system and its surroundings (Raymond, & John, 2010: 605). Heat energy transfer from one body to another is as a result of differences in temperatures of the two bodies. The process of heat energy transfer from one body to another is known as heat exchange. Various consequences such as temperature change occur as a result of this thermal energy exchange. The relationship between exchange of heat and temperature change in an object is dependent on specific heat capacity. Measurement of heat capacity is an investigation of entropy in a system and it is helpful in the investigation of basic thermodynamic properties of substances. Electrical method can be used to determine specific heat capacity of a solid and it involves heating the given solid of mass for a certain period of time and noting the temperature change for that period. The specific heat capacity can be obtained by relating various parameters such as voltage, current, time and mass. This technique will be used for this case and the process of heat exchange is applied. 1.1 Aims The main aims of this experiment were to; Assist in familiarization with various apparatus Determine specific heat capacity of a block Determine thermal conductivity of a rod 2.0 Theory 2.1 Specific Heat Capacity Heat capacity is of a particular substance is defined as the amount of energy needed to raise the temperature of a sample of the same substance by 10 C (Raymond, & John, 2010: 608). The SI unit for heat capacity is J/K. From the definition, heat energy Q raises the temperature by T. This can be represented as; Q = CT Specific heat capacity c of a substance is the heat capacity per unit mass (Raymond, & John, 2010: 608). Its SI unit is J/Kg°C. Considering the specific capacity of a substance, heating a unit mass of the same substance will result to corresponding temperature change. Heat Q or causing temperature change T or is given by Q = mcT or Where; m; mass of the substance c; specific heat capacity of the same substance in use being taken to be constant over temperature changeor . When a solid of mass is electrically heated from an initial temperature to a final temperature , for a certain length of time, a relationship exists between voltage and current supplied and these parameters. Power is energy transferred or work done per unit time. Therefore; power = work done (energy)/time But Electrical heat energy supplied or = Also, and for this case then the two equations can be equated; Specific heat capacity of this block can be calculated when temperature change, mass, voltage, current and time are known; Specific heat capacity 2.2 Thermal Conductivity Thermal conductivity is the amount of heat flowing per second through a unit area of plate of unit thickness when there is a temperature difference between the two faces of the plate (Roy, 2001: 153). Heat transfer occurs as a result of temperature difference in the two faces of the solid bar. Heat moving from high temperature face to low temperature face is therefore; 1. Directly proportional to the difference in temperature on the two faces of the bar. 2. Directly proportional to the cross-sectional area of the bar. 3. Directly proportional to time taken by heat to traverse between the faces. 4. Inversely proportional to inter-face difference. Therefore, this can be mathematically expressed as; Proportionality constant can be introduced for this case Where is the proportionality constant. This value is dependent on the material used to make that solid. This relation can be used to determine the heat flow (Watts) through the rod. 3.0 Apparatus Heat source (H1) – Comprises of a solid metal block with 4R7 resistor attached to it. A rod of unknown metal (M1) A heat sink (S1) 3.1 Part 1 3.1.1 Procedure Before the start of experiment, it was important to familiarize with the apparatus. 1. The experiment began with calculation of heat dissipation rate in Watts when voltage was set at 20V and the resistance of heating element was 4.7Ω. 2. The heating element was then connected to the power supply and both voltages were set to 20 volts. This was followed by accurate measurement of current through and voltage across the resistor. It was necessary to take note of these values because they were useful in the calculation of heat developed in the resistor. 3. The temperature rise of heated block (H1) was then recorded after every few seconds of heating. The process of recording temperature and corresponding time continued until the heating element turned off. 4. A graph of temperature versus time was drawn from the resultant values. The slope of the line of best fit was determined. 5. The result was then used in the calculation of specific heat capacity CH1 of the heated metal block H1. 3.1.2 Results and Discussions 1. Watts = Current × Voltage = Energy Change ÷ Change in time. Power (Watts) = V (volts) × I (Amperes) But V=IR therefore; I=V÷R Substituting this in the equationthen we shall have Time (s) Temperature (°C) 0 18 15 18 30 19 45 20 60 21 75 21 90 22 105 23 120 24 135 25 150 26 165 28 180 29 195 30 210 31 225 32 240 33 255 33 270 35 285 36 300 37 315 38 330 39 345 41 360 42 375 43 390 44 405 45 420 46 435 47 450 47 465 48 480 48 495 49 510 49 525 49 Function Therefore the gradient ()=0.0664 So as energy E increases with time thus the temperature increases with time; Differentiating E with respect to then we shall have Where; m is the mass CH1 is the specific heat capacity of the block H1. Mass of the block=1012g CH1 = 6.71968×10-2J/kg0C 3.2 Part 2 3.2.1 Procedure 1. The heating element was connected to power supply and both voltages were set to 23 volts. 2. The heated block (H1) was then allowed to heat at a steady state temperature up to 500C. 3. The equipment was assembled with metal rod M1 as follows; I. The source H1 II. The rod of unknown metal M1 III. Attach thermometers to both ends of the metal rod M1 IV. The heat sink S1 V. Attach a thermometer to the heat sink S1 4. The temperature rise of the heat sink S1 was recorded after every few seconds until the temperature of block S1 was above 350C. 5. A graph of temperature versus time was then drawn. 6. A tangent to the graph at a point P (250C) was then drawn and its gradient determined. Gradient was used to determine heat flow in Watts through the rod. 7. The results were used to predict the type of metal material in which the rod was made from. Results and Discussions Time (s) θ1 (°C) θ2 (°C) S1 (°C) 0 32 22 19 15 34 23 19 30 35 23 19 45 35 24 19 60 36 24 19 75 36 25 19 90 36 25 19 105 37 26 20 120 37 26 20 135 37 27 20 150 37 27 20 165 38 27 21 180 38 27 21 195 38 28 21 210 38 28 21 225 38 28 21 240 38 28 22 255 38 28 22 270 38 28 22 285 38 28 22 300 38 29 22 315 38 29 23 330 38 29 23 345 38 29 23 360 38 29 23 375 38 29 23 390 38 29 24 405 38 30 24 420 38 30 24 435 38 30 24 450 38 30 24 465 38 30 25 480 38 30 25 495 38 30 25 510 39 30 25 525 39 31 25 540 39 31 26 555 39 31 26 570 39 31 26 585 39 31 26 600 39 31 26 615 39 31 26 630 39 32 26 645 39 32 27 660 39 32 27 675 39 32 27 690 39 32 27 705 39 32 27 720 39 32 28 735 40 32 28 750 40 33 28 765 40 33 28 780 40 33 28 795 40 33 28 810 40 33 28 825 40 33 28 840 40 33 29 855 40 33 29 870 40 33 29 885 40 34 29 900 40 34 29 915 40 34 29 930 40 34 29 945 40 34 30 960 40 34 30 975 40 34 30 990 40 34 30 1005 40 34 30 1020 40 34 30 1035 40 34 31 1050 40 35 31 1065 40 35 31 1080 40 35 31 1095 41 35 31 1110 41 35 31 1125 41 35 31 1140 41 35 31 1155 41 35 31 1170 41 35 31 1185 41 35 32 1200 41 35 32 1215 41 36 32 1230 41 36 32 1245 41 36 32 1260 41 36 32 1275 41 36 32 1290 41 36 32 1305 41 36 32 1320 41 36 33 1335 41 36 33 1350 41 36 33 1365 41 36 33 1380 41 36 33 1395 41 36 33 1410 41 36 33 1425 41 36 33 1440 41 36 33 1455 41 36 33 1470 41 37 33 1485 41 37 33 1500 41 37 33 1515 41 37 34 1530 41 37 34 1545 42 37 34 1560 42 37 34 1575 42 37 34 1590 41 37 34 1605 41 37 34 1620 42 37 34 1635 42 37 34 1650 42 37 34 1665 42 38 35 Function of the tangential line is y=0.0119x+19.0538 Gradient=0.0119 Diameter of the rod =22.5mm, Length of the rod X=148mm=0.148M Cross-sectional area , V=23 volts, Resistance R of the heating element = 4.7 Ω. This material is likely to be made of lead with thermal conductivity of 34.7 Difference=37.2685-34.7=2.5685 The experimental value varies from the expected value due to errors incurred when conducting the experiment. Sources of Errors Sources of errors in the experiment included; Heat losses from the substance to the surrounding air Measurement errors that resulted from unequal distribution of heat Correction of Errors Errors could be avoided through several methods Heat losses could be avoided through lagging Equal heat distribution could be avoided by ensuring that the heating element covers all sides of the material used. References Raymond, A. S. & John W. J. Physics for Scientists and Engineers (6th Ed). California: Cengage Learning. 2010. Print. Roy, S. K. Thermal Physics and Statistical Mechanics. New Delhi: New Age International, 2001. Print. Read More

3. The temperature rise of heated block (H1) was then recorded after every few seconds of heating. The process of recording temperature and corresponding time continued until the heating element turned off. 4. A graph of temperature versus time was drawn from the resultant values. The slope of the line of best fit was determined. 5. The result was then used in the calculation of specific heat capacity CH1 of the heated metal block H1. 3.1.2 Results and Discussions 1. Watts = Current × Voltage = Energy Change ÷ Change in time.

Power (Watts) = V (volts) × I (Amperes) But V=IR therefore; I=V÷R Substituting this in the equationthen we shall have Time (s) Temperature (°C) 0 18 15 18 30 19 45 20 60 21 75 21 90 22 105 23 120 24 135 25 150 26 165 28 180 29 195 30 210 31 225 32 240 33 255 33 270 35 285 36 300 37 315 38 330 39 345 41 360 42 375 43 390 44 405 45 420 46 435 47 450 47 465 48 480 48 495 49 510 49 525 49 Function Therefore the gradient ()=0.0664 So as energy E increases with time thus the temperature increases with time; Differentiating E with respect to then we shall have Where; m is the mass CH1 is the specific heat capacity of the block H1.

Mass of the block=1012g CH1 = 6.71968×10-2J/kg0C 3.2 Part 2 3.2.1 Procedure 1. The heating element was connected to power supply and both voltages were set to 23 volts. 2. The heated block (H1) was then allowed to heat at a steady state temperature up to 500C. 3. The equipment was assembled with metal rod M1 as follows; I. The source H1 II. The rod of unknown metal M1 III. Attach thermometers to both ends of the metal rod M1 IV. The heat sink S1 V. Attach a thermometer to the heat sink S1 4.

The temperature rise of the heat sink S1 was recorded after every few seconds until the temperature of block S1 was above 350C. 5. A graph of temperature versus time was then drawn. 6. A tangent to the graph at a point P (250C) was then drawn and its gradient determined. Gradient was used to determine heat flow in Watts through the rod. 7. The results were used to predict the type of metal material in which the rod was made from. Results and Discussions Time (s) θ1 (°C) θ2 (°C) S1 (°C) 0 32 22 19 15 34 23 19 30 35 23 19 45 35 24 19 60 36 24 19 75 36 25 19 90 36 25 19 105 37 26 20 120 37 26 20 135 37 27 20 150 37 27 20 165 38 27 21 180 38 27 21 195 38 28 21 210 38 28 21 225 38 28 21 240 38 28 22 255 38 28 22 270 38 28 22 285 38 28 22 300 38 29 22 315 38 29 23 330 38 29 23 345 38 29 23 360 38 29 23 375 38 29 23 390 38 29 24 405 38 30 24 420 38 30 24 435 38 30 24 450 38 30 24 465 38 30 25 480 38 30 25 495 38 30 25 510 39 30 25 525 39 31 25 540 39 31 26 555 39 31 26 570 39 31 26 585 39 31 26 600 39 31 26 615 39 31 26 630 39 32 26 645 39 32 27 660 39 32 27 675 39 32 27 690 39 32 27 705 39 32 27 720 39 32 28 735 40 32 28 750 40 33 28 765 40 33 28 780 40 33 28 795 40 33 28 810 40 33 28 825 40 33 28 840 40 33 29 855 40 33 29 870 40 33 29 885 40 34 29 900 40 34 29 915 40 34 29 930 40 34 29 945 40 34 30 960 40 34 30 975 40 34 30 990 40 34 30 1005 40 34 30 1020 40 34 30 1035 40 34 31 1050 40 35 31 1065 40 35 31 1080 40 35 31 1095 41 35 31 1110 41 35 31 1125 41 35 31 1140 41 35 31 1155 41 35 31 1170 41 35 31 1185 41 35 32 1200 41 35 32 1215 41 36 32 1230 41 36 32 1245 41 36 32 1260 41 36 32 1275 41 36 32 1290 41 36 32 1305 41 36 32 1320 41 36 33 1335 41 36 33 1350 41 36 33 1365 41 36 33 1380 41 36 33 1395 41 36 33 1410 41 36 33 1425 41 36 33 1440 41 36 33 1455 41 36 33 1470 41 37 33 1485 41 37 33 1500 41 37 33 1515 41 37 34 1530 41 37 34 1545 42 37 34 1560 42 37 34 1575 42 37 34 1590 41 37 34 1605 41 37 34 1620 42 37 34 1635 42 37 34 1650 42 37 34 1665 42 38 35 Function of the tangential line is y=0.0119x+19.0538 Gradient=0.

0119 Diameter of the rod =22.5mm, Length of the rod X=148mm=0.148M Cross-sectional area , V=23 volts, Resistance R of the heating element = 4.7 Ω. This material is likely to be made of lead with thermal conductivity of 34.7 Difference=37.2685-34.7=2.5685 The experimental value varies from the expected value due to errors incurred when conducting the experiment.

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