Joule Tech Report by Charles Hunter

James Jouleâ&#x20AC;&#x2122;s Mechanical Equivalent of Heat Experiment Daniel Chapman, Emily Gore, Charles Hunter AME20213: Laboratory Exercise No. 1 Department of Aerospace and Mechanical Engineering University of Notre Dame February 25, 2010

Abstract The main purpose of this experiment was to determine the mechanical equivalent of heat by using a replica of James Jouleâ&#x20AC;&#x2122;s original experimental apparatus to convert the mechanical energy of falling weights into thermal energy in a thermally isolated water container. By accounting for possible avenues for energy losses, the experimental specific heat of water was determined to within 4.23 % of the accepted value of 4186 kJ/kg K. The accepted value was then used to predict the change in temperature. The computer model generated was capable of predicting the final temperature after 100 drops to within 0.7 % of the experimentally observed value.

Contents 1 Nomenclature …………………………………………………………………………………...... 4 2 Introduction ……………………………………………………………………………………….. 5 3 Experimental Approach and Procedure …………………………………………… 7 4 Results ……………………………………………………………………………………………….. 8 4.1 Comparison of Thermistor to Thermocouple…………………………………….. 8 4.2 Calculating Specific Heat ……………………………………………………………….. 9 4.3 Energy Model Comparison ……………………………………………………………..…….. 10 4.4 Model with Liquid Mercury ………………………………………………………………….. 10

5 Discussion …...…………………………………………………………………………………… 11 6 Conclusions …………………………………………………………………………….……….. 12 7 References …………………………………………………………………………….……........ 13 Appendices Appendix A: Answers to Application Problems ……………………….…………… 14 Appendix B: MATLAB Code ………………………………………..……………………………..… 18 Appendix C: Uncertainty Analysis ……………………………………………………………… 23 2

Figures Figure 1: Replica of the Joule Apparatus

Figure 2: Flow Chart for Temperature Prediction Computer Codes

Figure 3: Flow Chart for Experimental Specific Heat Codes

8 9

Figure 4: Number of Drops vs. Temperature Figure 5: Rise in Temperature vs. Number of Weight Drops Figure 6: Linear Regression Analysis of Thermocouple Data

11 15

1 Nomenclature English Specific Heat (J/Kg-K) Energy (J) Height (m) Moment of Inertia (Kg-m2) Kinetic Energy (J) Mass (kg) Temperature (K) Velocity (m/s) Volume (m3)

Greek Density (Kg/m3) Radial Velocity (rad/s) Change in

Subscripts Mechanical Liquid Pressure Thermal

2 Introduction The purpose of this laboratory experiment was to quantify the conversion of mechanical energy to heat energy by measuring the mechanical energy expended and heat energy gained, similar in nature to James Joule’s heat experiment conducted circa 1850. Using a model to predict actual energy losses would be formulated by taking into account energy lost to different aspects of the experimental conditions. Joule performed a series of experiments between 1845 and 1847 to obtain a value which proved the existence of an equivalent relation between force and heat. His experiment had been carefully designed to minimize any energy losses. Some losses did occur, however, and Joule extensively predicted and quantified the possible losses. A quantitative value for the mechanical force equivalent to produce a certain amount of heat had first been presented by James Joule in 1850. To aid Joule, he relied on the works of previous scientists such as Count Rumford, Sir Humphry Davy, Faraday, and Mayer. His experimental results established that a known amount of work would generate a proportional amount of heat. The value Joule calculated to increase the temperature of a pound of water by 1 degree Fahrenheit is 772 ft-lbf. The accepted value today in the English Engineering system is 778 ft-lbf which only differs from Joule’s by less than 0.8 %. In S.I. units the specific heat value is 4186 J/kg-K. A description of Joule’s apparatus follows. Two known weights were connected to a pulley system and dropped from a known height, turning the side pulleys. The center pulley rotated a paddle wheel with multiple fins inside of a thermally insulated container filled with distilled water. The rotation agitated the water and caused the temperature of the liquid to rise. While one weight dropped, the wire on the other side wound up. The second weight was then added and dropped in the same manner as the first weight. When this process was repeated several times, a measurable increase in the water’s temperature was observed. A picture of the full-scale replica with its designer, Leon Hluchota, is shown in Figure 1.

Figure 1: Replica of the Joule Apparatus In order to measure the increase of the waterâ&#x20AC;&#x2122;s temperature and resistance, a type-K thermocouple and a thermistor (Omega Model No. 44033) were mounted inside the insulated container so they were exposed to the water. A digital thermocouple indicator and a digital ohmmeter located outside of the container reported the measured resistance of the thermistor and temperature of the water. The energy gained from dropping the weight was transferred to the paddle wheel which then became the gain in the heat energy of the water, as shown in following equation, [1] where is the weight, is the height which the weight is dropped, is the change in mechanical energy, is the change in thermal energy, is the mass of the liquid, is the specific heat at constant pressure of the liquid, and is the increase in the liquidâ&#x20AC;&#x2122;s temperature. The mass of the water can be determined from the following equation, [2] where

is the density of the liquid and

is the volume of the liquid.

James Joule originally conducted five series of this experiment using water, mercury, and cast iron. The experiment with water involved over one thousand trials lasting about an hour each. In the recreation of his experiment only one series using water was carried out, and involved one-hundred trials (drops) lasting for a duration of approximately 45 minutes.

3 Experimental Approach and Procedure Before beginning the experiment, the weight drop distance was recorded as 1.42 m, ambient temperature as 297.87 K, thermistor resistance as 2.28 kâ&#x201E;Ś, and start time as 0.00 s. For the first weight drop the amount of time it took to fall to the ground was recorded as 23.0 s. This value was used to calculate the final velocity, which was assumed to be twice the average velocity, per simple kinematics. Once the weight hit the floor it was removed and another weight was added to the opposite-side wire. The weight was allowed to drop and the temperature, time, and resistance was recorded after 10 consecutive weight drops for 100 drops. In complement with the outputs from the thermocouple and thermistor, it was then possible to determine the work from each drop and the corresponding gain in thermal energy. As a means of greater efficiency, several MATLAB codes were written to aid in the calculations of the experimental estimate of specific heat, as well as different models to predict the change in temperature and the final temperature after 100 drops. The first pair of codes, joule_predict2.m and joule_no_loss_predict.m (both to be found in Appendix B: MATLAB Codes), were used in conjunction with the accepted value for the specific heat in order to predict the temperature change over 100 drops. A flow chart of these two models is laid out below in Figure 2. Convert starting temperature and other measurands to SI units.

Calculate energy from single drop

Calculate magnitude of energy lost in transformation to thermal energy (if using modified model)

Use loop to simulate 100 drops, subtracting lost energy each time (if using modified model)

Use cumulative sum of all iterations in concjunction with accepted value of specific heat of water to find temperature rise with each weight drop

Figure 2: Flow Chart for Temperature Prediction Computer Codes. The second pair of codes, spec_heat3.m and no_loss_spec_heat3.m (both to be found in Appendix B: MATLAB Codes), were used to calculate the specific heat from the experimental data. A flow chart of these models is laid out below in Figure 3.

Convert measurands to SI units

Find sources of energy loss (for modified model, if using simple model, ignore these)

Find work done with each drop

Run 100 iterations to simulate 100 drops

Calculate specific heat and percent error compared to accepted value Figure 3: Flow Chart for Experimental Specific Heat Codes. After the codes had been used to find the experimental estimation of specific heat and predict the final temperature in relation to the generally accepted value for specific heat, uncertainty was calculated for each result and was plotted and analyzed using MATLAB software.

4 Results 4.1 Comparison of Thermistor to Thermocouple As seen in Figure 4, the thermistor resistance reading (converted to temperature using an equation found in [3]), readings from the thermocouple and the prediction temperature values from the model are compared. All three data sets have a similar positive linear trend. Using the data from the model, a linear best fit curve was found in order to compare the data visually.

Figure 4: Number of Drops vs. Temperature The temperature values between the thermistor and thermocouple readings differed by an average of 0.013 %. This small discrepancy makes sense because both instruments were placed directly into the liquid during the experiment. The thermistor was used primarily for validational purposes. The difference between the thermistor reading and the model values were an average of 0.009 %, and the thermocouple varied from the model by an average of 0.008 %. These minuscule percent differences are due to the level of accuracy of the generated model. An error was calculated for each member of the data set. The thermistor error of 0.110 was based on a zeroth degree of uncertainty of reading the value from the meter based on its resolution, as well as the instrumental uncertainty associated with the device. The thermocouple error was only calculated using the zeroth uncertainty which was found to be 0.500. Finally the model's uncertainty of 0.020 was found using the uncertainty analysis seen in Appendix C. 4.2 Calculating Specific Heat Using a modified model in which assumed energy losses due to hitting the ground and the starting and stopping of the rotation of the pulleys and paddle wheel (explained in problem 1-3), the experimental value for the specific heat of water was found to be 4363.2 J/kg K. This differs from the accepted value of 4186 J/kg K [2] by 4.23 %.

4.3 Energy Model Comparison Using a model in which it is assumed that there are no energy losses involved in the transformation of energy from the dropping of the weights into thermal energy in the fluid, an overall temperature rise of 0.719 ± 0.1 K was predicted, which differs from the experimental value of 0.611 ± 0.1 K by 17.72 %. This error is particularly high because there are a number of locations where energy is lost and therefore not transformed entirely into thermal energy. This model still did, however, predict a final temperature of 298.60 ± 0.1 K, which fell within 0.70 % of the experimental value. As a point of clarification, percent error was calculated with reference to the Celsius scale, as use of the Kelvin scale led to unrealistically low estimates of the error. Using the modified model, a temperature change of 0.621 ± 0.1 K was predicted, which differed from the experimental value by only 2.78 %. This error value appears high, but is important to note that the model also predicted the final temperature to be 298.51 ± 0.1 K as opposed to the experimental value of 298.43 ± 0.1 K. This relation demonstrates only a 0.2 % error, a marked improvement over the “simple” model. 4.4 Model with Liquid Mercury The MATLAB code joule_predict2.m (Appendix B: MATLAB Code) used to predict the temperature rise for water was also altered into joule_predict_Mercury.m (to be found in Appendix B: MATLAB Code), which predicts the temperature rise for an equivalent volume of liquid mercury. This was done by switching the specific heat of water to that of liquid mercury, 140 J/kg-K [2], and changing the density of water to that of liquid mercury, 0.0135 kg/mL [2]. The predicted temperature rise for an equivalent volume of liquid mercury for the same amount of work done is shown in Figure 5. The temperature rise in the liquid mercury was predicted to be 1.24 K from the starting value. The final temperature prediction of mercury using the model was found to be 0.203 % higher than the final model prediction of water. It can thus be concluded that mercury is more efficient than water in converting mechanical energy into thermal energy.

Figure 5: Rise in Temperature vs. Number of Weight Drops

5 Discussion The notable value of the percent error of the model for specific heat can be attributed to the very sensitive nature of the specific heat. Due to the relatively small number of drops performed in this experiment (only 100 drops were performed as compared to Jouleâ&#x20AC;&#x2122;s thousands of iterations), it was difficult to obtain a very precise value for the temperature change. Because a relatively large amount of energy to produce a relatively small effect on the temperature, the resolution of the thermocouple led to a high degree of uncertainty in relation to the calculation of the specific heat. This may be rectified either with a higher number of iterations (which would give a broader range over which to apply the calculations) or by improving the resolution or sensitivity of the temperature measurement devices. Observation of the small percent error for the no-loss model suggests that the measurement of the specific heat is much more sensitive than the prediction of temperature change because of the large amount of energy necessary to raise the temperature of the water by a relatively small increment. As mentioned in Section 4.4, the lower specific heat of mercury makes it ostensibly more â&#x20AC;&#x153;efficientâ&#x20AC;? in accepting mechanical energy and converting it into thermal energy and, subsequently, a rise in temperature. The modified model, to be discussed further below, demonstrated a still better approximation of the final temperature. 11

In light of the apparent inaccuracy of the original temperature prediction code, it was necessary to explore ways in which energy may have been lost to other processes. One such assumed process was the energy lost due to the weights hitting the ground, which occurred with each drop. This was approximated by assuming that all of the kinetic energy of the weight was transferred to the ground upon contact, using the formula [3] where KE is the kinetic energy in Joules, m is the mass in kilograms, and v is the velocity in meters per second. The velocity was approximated as twice the average velocity, obtained through the original measurement of the drop height and the time it takes to traverse that distance. Another possible location of energy loss was in the starting and stopping of rotation of each of the pulleys. This was approximated by finding the rotational kinetic energy, given by the formula [4] where is the moment of inertia about the rotating axis and ω is the angular velocity in radians per second. This was used for the wood and aluminum outer pulleys in addition to the center pulley and the rotating members of the brass stirring apparatus. The paddles of the brass apparatus was modeled as three plates rotating their edges. When these formulas were used in conjunction with known measurands such as the mass and density of each of the components involved, a second MATLAB model, spec_heat3.m (Appendix B: MATLAB Codes) was constructed to more accurately model the transformation of mechanical energy into thermal energy.

6 Conclusions Using specheat3.m, the experimental value for the specific heat of water was found to be 4363.2 J/kg K. This differed from the accepted value of 4186 J/kg K [2] by 4.23 %. Using joule_predict2.m, a temperature change of 0.628 ± 0.1 K was predicted, which differed from the experimental value by only 2.78 %. Just as Joule was forced to account for energy losses outside of his control, the final computer model used took into account mechanical energy lost in the rotational kinetic energy of the pulleys and paddle wheel, as well as thermal energy lost to the brass components of the sealed water container. The temperature prediction model also predicted the final temperature to be 298.51 ± 0.1 K as opposed to the experimental value of 298.43 ± 0.1 K, which demonstrates only a 0.2 % error. The larger percent error associated with the specific heat in comparison to that of the temperature prediction reinforces the sensitive nature of energy conversion between thermal and mechanical energy, with a great amount of mechanical energy to produce a relatively minuscule effect on temperature. This problem was rectified by Joule with a greater number of iterations in order to gather data over a broad range, something that was outside the time constraints for this experiment.

7 References [1] “Mercury Element Facts.” Chemicool. 24 Feb. 2010 <http://www.chemicool.com/elements/mercury.html> [2] “Specific Heat Capacity.” Wikipedia. 22 Feb. 2010. 24 Feb. 2010 <http://en.wikipedia.org/wiki/Specific_heat_capacity> [3] “Thermistor Elements.” Omega. 17 Feb. 2010 <http://www.nd.edu/~pdunn/www.ame250/therm>

Appendices Appendix A: Answers to Application Problems (Figures and equations shown in the body of the report are referred to but not reproduced here) Problem 1-1: The thermistor resistance readings, that were obtained experimentally were converted to temperatures by using the equation ( )

( ) ,

(3)

where and are constants with values 1.468*10^-3, 2.383*10^-4 and 1.007*10^-7 respectively. The readings from the thermocouple, converted thermistor values and the model prediction temperature values are compared in Figure 4. All three data sets have a similar positive linear trend. Using the data from the model, a linear best fit curve was found with the equation, , in order to compare the data visually. The temperature values between the thermistor and thermocouple readings differed by an average of 0.013 %. This small discrepancy would make sense because both instruments were placed directly into the liquid during the experiment. The thermistor was used primarily for validational purposes. The difference between the thermistor reading and the model values were an average of 0.009 % and the thermocouple varied from the model by an average of 0.008 %. These minuscule percent differences are due to the accuracy of our model. An error was calculated for each of the data set. The thermistor error of 0.110 was based on a zeroth uncertainty of reading the value from the meter and the instrumental uncertainty associated with the device. The thermocouple error was only calculated using the zeroth uncertainty which was found to be 0.500. Finally the model's uncertainty of 0.020 was found using the uncertainty analysis seen in Appendix C. Using a modified version of jcaleyIII.m, a least squares regression analysis was applied to the thermocouple data with 95.0 % confidence and 2 as the student value as seen in Figure 6. The major results can be found at the top of the figure including the sum of the product of x and y values , and The data clearly fits in between the modeling uncertainty curves, and the theoretical curve seems to be a best fit line.

Figure 6: Linear Regression Analysis of Thermocouple Data Problem 1-2: Using a modified model in which assumed energy losses due to hitting the ground and the starting and stopping of the rotation of the pulleys and paddle wheel (explained in problem 1-3), the experimental value for the specific heat of water was found to be 4363.2 J/kg K. This differs from the accepted value of 4186 J/kg K by 4.23 %. This value of error can be attributed to the very sensitive nature of the specific heat. Due to the relatively small number of drops performed in this experiment (only 100 drops were performed as compared to Jouleâ&#x20AC;&#x2122;s thousands of iterations), it was difficult to obtain a very precise value for the temperature change. Because a relatively large amount of energy to produce a relatively small effect on the temperature, the resolution of the thermocouple led to a high degree of uncertainty in relation to the calculation of the specific heat. This may be rectified either with a higher number of iterations (which would give a broader range over which to apply the calculations) or by improving the resolution or sensitivity of the temperature measurement devices. Problem 1-3: Using a model in which it is assumed that there are no energy losses involved in the transformation of energy from the dropping of the weights into thermal energy in the fluid, 15

an overall temperature rise of 0.7194 ± 0.1 K was predicted, which differs from the experimental value of 0.611 ± 0.1 K by 17.72 %. This error is particularly high because there are a number of locations where energy is lost and therefore not transformed entirely into thermal energy. This model still did, however, predict a final temperature of 298.5989 ± 0.1 K, which falls within 0.7 % of the experimental value. This observation suggests that the measurement of the specific heat is much more sensitive than the prediction of temperature change because of the large amount of energy necessary to raise the temperature of the water by a relatively small increment. In light of the apparent inaccuracy of the original temperature prediction code, it was necessary to explore ways in which energy may have been lost to other processes. One such assumed process was the energy lost due to the weights hitting the ground found to be 0.110 J, which occurred with each drop. This was approximated by assuming that all of the kinetic energy of the weight was transferred to the ground upon contact, using the formula (1) where KE is the kinetic energy in Joules, m is the mass in kilograms, and v is the velocity in meters per second. The velocity was approximated as being constant throughout the drop. Another possible location of energy loss was in the starting and stopping of rotation of each of the outer pulleys found to be 0.539 J. This was approximated by finding the rotational kinetic energy, given by the formula (2) where I is the moment of inertia about the rotating axis and ω is the angular velocity in radians per second. This was used for the wood and aluminum outer pulleys in addition to the center pulley and the rotating members of the brass stirring apparatus whose loss was found as 3.965 J. The paddles of the brass apparatus was modeled as a plate rotating about one of its edges. When these formulas were used in conjunction with known measurands such as the mass and density of each of the components, a second MATLAB model was constructed to more accurately model the transformation of mechanical energy into thermal energy. Using the modified model, a temperature change of 0.6281 ± 0.1 K was predicted, which differs from the experimental value by only 2.78 %. This error value appears high, but is important to note that the model also predicted the final temperature to be 298.5067 ± 0.1 K as opposed to the experimental value of 298.4278 ±0.1 K. This relation demonstrates only a 0.2 % error, a marked improvement over the “simple” model. Problem 1-4: The MATLAB code joule_predict_Mercury.m (Appendix B: MATLAB Code) predicts the temperature rise for an equivalent volume of liquid mercury. This was done by switching the specific heat of water to that of liquid mercury, 140 J/Kg-K, and changing the density of water to that of liquid mercury, 0.013 534 Kg/mL in joule_predict.m (Appendix B: MATLAB Code). The predicted temperature rise for an equivalent volume of liquid 16

mercury for the same amount of work done is shown in Figure 5. The temperature rise in the liquid mercury was predicted to be 1.24 K, starting from 297.87 K and rising to 299.11 K. The final temperature prediction of mercury using the model was found to be 0.203 % higher than the final model prediction of water. It can be concluded that mercury is more efficient than water in converting mechanical energy into thermal energy.

Appendix B: MATLAB Code spec_heat3.m Code used for application problem 1-2 % Danny Chapman % AME 20213 % Lab 1: Joule Experiment % 22 February 2010 % spec_heat3.m calculates experimental value of specific heat for water. clear all; c_theory = 4186; % Convert thermistor values to Kelvin. f = [76.5 76.6 76.7 76.8 76.9 77.0 77.1 77.3 77.4 77.5 77.6]; K = (f - 32) * (5/9) + 273.15; % Convert drop height to meters. height_inches = 55; h_m = 2.54 * height_inches / 100; % Find average weight dropped in Newtons. weight_1 = 139.972; weight_2 = 141.662; weight_avg = (weight_1 + weight_2) * 0.5; % Account for energy lost due to weights hitting the ground, assuming the % kinetic energy at impact is all imparted to the ground. It was assumed % that the final velocity was twice the average velocity. v_weight = .06184; E_hit = 0.5 * (weight_avg / 9.81) * (2 * v_weight)^2; % Account for energy lost due to starting and stopping rotation of pulleys % using moments of inertia for the paddle wheel, center pulley, and outer % pulleys. omega_out = 2.4348; m_pine = 1.939; r_pine = .1524; I_pine = 0.5 * m_pine * (r_pine)^2; m_al = .6255; r_al = .0254; I_al = 0.5 * m_al * (r_al)^2; E_outer = 0.5 * (I_pine + I_al) * (2 * omega_out)^2 * 2; omega_paddle = 2 * 7.3044; m_paddle = 1.200; r_paddle = .1524; I_paddle = (1/12) * m_paddle * (r_paddle)^2 + m_paddle * (r_paddle/2)^2; E_paddle = 0.5 * I_paddle * (2 * omega_paddle)^2; % Calculate work done [J] with each drop. work_drop = weight_avg * h_m; for i = 1:100; work_done(i,1) = i*work_drop - 2*i*E_outer - i*E_hit - 6*i*E_paddle; end % Calculate mass of water [kg] from known density and volume. volume_mL = 5830; density_kg_mL = .001; mass_L = volume_mL * density_kg_mL; % Use Joule's equation E = mc(dT) to find experimental specific heat c_P. E_therm = work_done(100,1); d_T = K(11) - K(1)

mass_brass = 1.378 + 1.2 * 3 + 2.037; c_brass = 380; c_P = (E_therm - mass_brass * c_brass * d_T) / (mass_L * d_T); % Print outputs. disp('The experimental specific heat is'); disp(c_P); disp('J/kg K'); disp('This differs from the theoretical value by'); error = (c_P - c_theory) / (c_theory) * 100; disp(error); disp('%'); %d_TH = d_T + .1; %c_PH = (E_therm - mass_brass * c_brass * d_TH) / (mass_L * d_TH); %disp('The higher boundary of uncertainty due to temperature is: '); %disp(c_PH); %errH = (c_PH - c_theory) / (c_theory) * 100; %disp(errH);

no_loss_spec_heat3.m Code used for application problem 1-2 % AME 20213 % Lab 1: Joule Experiment % 22 February 2010 % no_loss_spec_heat3.m calculates experimental value of specific heat for water. clear all; c_theory = 4186; % Convert thermistor values to Kelvin. f = [76.5 76.6 76.7 76.8 76.9 77.0 77.1 77.3 77.4 77.5 77.6]; K = (f - 32) * (5/9) + 273.15; % Convert drop height to meters. height_inches = 55; h_m = 2.54 * height_inches / 100; % Find average weight dropped in Newtons. weight_1 = 139.972; weight_2 = 141.662; weight_avg = (weight_1 + weight_2) * 0.5; % Calculate work done [J] with each drop. work_drop = weight_avg * h_m; for i = 1:100; work_done(i,1) = i*work_drop; end % Calculate mass of water [kg] from known density and volume. volume_mL = 5830; density_kg_mL = .001; mass_L = volume_mL * density_kg_mL; % Use Joule's equation E = mc(dT) to find experimental specific heat c_P. E_therm = work_done(100,1); d_T = K(11) - K(1); mass_brass = 1.378 + 1.2 * 3 + 2.037; c_brass = 380; c_P = (E_therm - mass_brass * c_brass * d_T) / (mass_L * d_T); % Print outputs. disp('The experimental specific heat is'); disp(c_P); disp('J/kg K');

disp('This differs from the theoretical value by'); error = (c_P - c_theory) / (c_theory) * 100; disp(error); disp('%');

joule_no_loss_predict2.m Code used for application problem 1-3 % Measurements and Data Analysis % Joule Experiment % 22 February 2010 % % % %

joule_no_loss_predict2.m is a program that predicts the temperature rise over a 100-drop iteration of Joule's experiment to determine the mechanical equivalent of heat. This program does NOT take into account energy losses due to rotational momentum, hitting the ground, etc. clear all; % Joule's calculated mechanical equivalent of heat, J/kg K: c_J = 4186; % Record a starting temperature value: T_o = input('Enter a starting value for the temperature in Celsius: '); % Energy from single drop, taking into account ONLY the weight drop: weight_1 = 139.972; weight_2 = 141.662; weight_avg = (weight_1 + weight_2) * 0.5; height_inches = 55; h_m = 2.54 * height_inches / 100; work_per_drop = h_m * weight_avg; % All energy is assumed to be transferred to thermal energy: E_thermal = work_per_drop; % Calculate mass of water [kg] from known density and volume, and take into % account the properties of the brass casing and paddle: volume_mL = 5830; density_kg_mL = .001; mass_L = volume_mL * density_kg_mL; mass_brass = 1.378 + 1.2 * 3 + 2.037; c_brass = 380; % Use loop to predict temperatures over 100 drops: for j = 1:100; T(j,1) = T_o + j * (E_thermal) / (mass_L * c_J + mass_brass * c_brass); end T = T + 273; for f = 1 % If outputs desired in Farenheit: for k = 1:100; F(k,1) = T(k,1) * 9/5 + 32; end

joule_predict2.m Code used for application problem 1-3 % Measurements and Data Analysis % Joule Experiment % 22 February 2010 % joule_predict2.m is a program that predicts the temperature rise over a % 100-drop iteration of Joule's experiment to determine the mechanical

% equivalent of heat. clear all; % Joule's calculated mechanical equivalent of heat, J/kg K: c_J = 4186; % Record a starting temperature value: T_oF = 76.5; T_o = (T_oF - 32) * (5/9) + 273.15; % Energy from single drop, taking into account ONLY the weight drop: weight_1 = 139.972; weight_2 = 141.662; weight_avg = (weight_1 + weight_2) * 0.5; height_inches = 55; h_m = 2.54 * height_inches / 100; work_per_drop = h_m * weight_avg; % Now take into account energy lost to starting and stopping rotation of % the pulleys: omega_out = 2.4348; m_pine = 1.939; r_pine = .1524; I_pine = 0.5 * m_pine * (r_pine)^2; m_al = .6255; r_al = .0254; I_al = 0.5 * m_al * (r_al)^2; E_outer = 0.5 * (I_pine + I_al) * (2 * omega_out)^2 * 2; omega_paddle = 2 * 7.3044; m_paddle = 1.200; r_paddle = .1524; I_paddle = (1/12) * m_paddle * (r_paddle)^2 + m_paddle * (r_paddle/2)^2; E_paddle = 0.5 * I_paddle * (2 * omega_paddle)^2; % Take into account energy lost due to weight hitting ground: v_weight = .06184; E_hit = 0.5 * (weight_avg / 9.81) * (2 * v_weight)^2; % Determine fraction of energy converted to thermal energy, assuming there % are no other energy losses: E_thermal = work_per_drop - 2*E_outer - E_hit - 6*E_paddle; % Calculate mass of water [kg] from known density and volume, and take into % account the properties of the brass casing and paddle: volume_mL = 5830; density_kg_mL = .001; mass_L = volume_mL * density_kg_mL; mass_brass = 1.378 + 1.2 * 3 + 2.037; c_brass = 380; % Use loop to predict temperatures over 100 drops: for j = 1:100; T(j,1) = T_o + j * (E_thermal) / (mass_L * c_J + mass_brass * c_brass); end for k = 1:100; F(k,1) = T(k,1) * 9/5 + 32; end T(100) - T(1)

joule_predict_Mercury.m Code used for application problem 1-4 % Measurements and Data Analysis % Joule Experiment % 22 February 2010

% joule_predict2.m is a program that predicts the temperature rise over a % 100-drop iteration of Joule's experiment to determine the mechanical % equivalent of heat. clear all; % Joule's calculated mechanical equivalent of heat, J/kg K: c_J = 140; % Record a starting temperature value: T_oF = 76.5; T_o = (T_oF - 32) * (5/9) + 273.15; % Energy from single drop, taking into account ONLY the weight drop: weight_1 = 139.972; weight_2 = 141.662; weight_avg = (weight_1 + weight_2) * 0.5; height_inches = 55; h_m = 2.54 * height_inches / 100; work_per_drop = h_m * weight_avg; % Now take into account energy lost to starting and stopping rotation of % the pulleys: omega_out = 2.4348; m_pine = 1.939; r_pine = .1524; I_pine = 0.5 * m_pine * (r_pine)^2; m_al = .6255; r_al = .0254; I_al = 0.5 * m_al * (r_al)^2; E_outer = 0.5 * (I_pine + I_al) * (2 * omega_out)^2 * 2; omega_paddle = 2 * 7.3044; m_paddle = 1.200; r_paddle = .1524; I_paddle = (1/12) * m_paddle * (r_paddle)^2 + m_paddle * (r_paddle/2)^2; E_paddle = 0.5 * I_paddle * (2 * omega_paddle)^2; % Take into account energy lost due to weight hitting ground: v_weight = .06184; E_hit = 0.5 * (weight_avg / 9.81) * (2 * v_weight)^2; % Determine fraction of energy converted to thermal energy, assuming there % are no other energy losses: E_thermal = work_per_drop - 2*E_outer - E_hit - 6*E_paddle; % Calculate mass of water [kg] from known density and volume, and take into % account the properties of the brass casing and paddle: volume_mL = 5830; density_kg_mL = .013534; mass_L = volume_mL * density_kg_mL; mass_brass = 1.378 + 1.2 * 3 + 2.037; c_brass = 380; % Use loop to predict temperatures over 100 drops: for j = 1:100; T(j,1) = T_o + j * (E_thermal) / (mass_L * c_J + mass_brass * c_brass); end for k = 1:100; F(k,1) = T(k,1) * 9/5 + 32; end plot(T,'-') xlabel('Number of Drops') ylabel('Temperature [K]') title('Expected Temperature Rise for Liquid Mercury') T(100) - T(1)