final yes 05_Enthalpy_PostLab_Spr24

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Feb 20, 2024

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Enthalpy and Specific Heat Name: Arnesha Jackson Date of Work Feb. 7th 2024 TOTAL: 20 points
Arnesha Jackson Enthalpy and Specific Heat Page of 2 8 CHM 113 POST-LAB Enthalpy and Specific Heat 1. Insert ONE picture of yourself in full PPE here ( include the Styrofoam cup ). **Remember to (1) show your full body so that we see you are wearing shoes; (2) wear your safety glasses and gloves; (3) cover your lower legs (socks are required, even in Arizona); (4) tie back long hair in a ponytail or a bun; (5) remove jewelry.
Arnesha Jackson Enthalpy and Specific Heat Page of 3 8 2. Complete Table 1 with your experimental data for the masses of the water and the unknown metal strip in Part 1 . 3. Complete Table 2 with your experimental data for the temperature changes for the water in your calorimeter in Part 1 , and calculate the temperature change ( Δ T) for each trial. Show your work for calculating Δ T below the table and include units. Calculations of Δ T water Trial 1: 21.8C-17.4C = 4.4C Trial 2: 23.1C- 18.2C = 4.9C Trial 3: 21.9C - 18.4C = 3.5C Table 1: Mass Data Mass (g) Trial 1 Trial 2 Trial 3 Water 49.9g 50.1g 49.5g Metal Strip 23.6g 23.6g 23.6g T = T final T initial Table 2: Specific Heat Data Time (minutes) Calorimeter Water Temperature (°C) Trial 1 Trial 2 Trial 3 Initial 17.4 C 18.2 18.4 1 minute 22.4 C 22.4 21.5 2 minutes 22.0 C 22.4 21.6 3 minutes 21.9 C 22.7 21.7 4 minutes 21.9 C 22.9 21.8 5 minutes 21.8 C 23.1 21.9 Δ T water 4.4 C 4.9 C 3.5 C
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Arnesha Jackson Enthalpy and Specific Heat Page of 4 8 4. Answer the following: Hint: Refer to the equations in the Introduction segment of the Procedure document. a. Calculate Δ T metal for each of the three trials. Show your work and include units: Hint: Table 2 shows the Δ T for the water. For the Δ T of your metal, consider the initial temperature of the metal (Part 1, Step 10 in Procedure) as well as the final temperature once the calorimeter contents reached thermal equilibrium. Calculations of Δ T metal Trial 1: 21.8C - 92.0C = -70.2 C Trial 2: 22.9C- 92.0 = -69.1.0C Trial 3: 22.2C- 92.0C = -69.8 b. Calculate the amount of heat absorbed by the water ( q water ) in each trial. Show your work and include units. Calculations of heat absorbed by water ( q water ) Trial 1: (49.9g)(4.184 J/s C)(4.4C) = 919 J Trial 2: (50.1g)(4.184 J/s C)(4.9C) = 1027 J Trial 3: (49.5g)(4.184 J/s C)(3.5C) = 725 J c. Determine the amount of heat released by the unknown metal ( q metal ) in each trial. Show your work/logic. Heat released by metal ( q metal ) Trial 1: (23.6g)(-70.2C)(0.554 J/g C) = -918 J Trial 2: (23.6g)(-70.0C)(0.622 J/g C) = -1028 J Trial 3: (23.6g)(-69.8C)(0.440 J/g C) = -725 J
Arnesha Jackson Enthalpy and Specific Heat Page of 5 8 d. Calculate the specific heat ( c or c sp ) of the unknown metal for each trial. Show your work and include units. Calculations of Specific Heat Trial 1: (23.6g) (x)(-70.2C) = -918J, x= 0.554 J/g C Trial 2: (23.6g)(x)(-70.0C) = -1028 J x=0.622 J/g C Trial 3:(23.6g)(x)(-69.8C) = -725 J x= 0.440 J/g C e. What is the average specific heat for all three trials? Show your work and include units. 5. Putting it all together! a. Using the table of Specific Heat Capacities in the Pre-Lab Introductory Information Handout, compare the average specific heat capacity you calculated in Question 4 to the values in the table to determine the identity of the unknown metal. Explain, in 2 – 3 sentences, how you know. b. If you repeated this experiment with a strip of silver with the same mass as the strip of metal you used in this experiment, would the water in the calorimeter heat up to a higher or lower temperature using silver than it did for your experimental metal? Explain, in 2 – 3 sentences, your reasoning. Hint: Compare the specific heat value of Ag(silver) with that of your experimental metal. Average Specific Heat: (0.554 J/g C + 0.622 J/g C + 0.440 J/g C) /3 = 0.539 J/ g C Unknown metal _________NICKLE_____________________ The specific heat indicated that the numbers were closely related to that of Nickel, and nickel is decently soft metal. The closest metal to my calculated specific heat of 0.539 J/g C is nickel. Nickel has a specific heat of 0.502. The difference in 0.037 J/g C could be due to the unequal heat distribution across metal, unknown particles could’ve been attached to the metal plates, or an error in initial weighing.
Arnesha Jackson Enthalpy and Specific Heat Page of 6 8 6. Complete Combined Table 3 & 4 with your experimental data for the cold pack and hand warmer experiments ( Part 2 ) below. Highlight (in yellow) the minimum temperature for the cold pack and the maximum temperature for the hand warmer . 7. Plot your data from Tables 3 and 4 (TWO data sets from Question 6) on one x-y scatter plot using a graphing program. Your graph should show the temperature as a function of time for TWO sets of data on ONE graph. You will not receive credit if you draw your graph by hand. Excel or Google Sheets are good choices that you can learn how to use quickly if you don’t already have a favorite graphing program. On your graph, the x-axis should be the time (in seconds) , and the y-axis should be the temperature (in °C) at that time. Make sure to label the axes including the units in parentheses but REMOVE the chart title and any default legend put in by the graphing software. Also remove grid lines and labels on individual data points. Silver has a specific heat capacity of 0.240 J/g C which is lower than nickels’ specific heat capacity of 0.502 J/g C. lower temps are expected with the calorimeter water since it takes less heat to warm the metal than it would transfer a lower amount of heat to the water. Tables 3 & 4 Cold Pack Hand Warmer Time (sec) Temperature ( ° C) Temperature ( ° C) Initial 22.9C 29.5C 30 1 C 50 C 60 -1.2 C 55.5 C 90 -1.8 C 66 C 120 -1.4 C 75.8 C 150 -1.1 C 80.9 C 180 0.4 C 82.2 C 210 0.5 C 82.5 C 240 0.6 C 82.7 C 270 0.8 C 82 C 300 1.2 C 81.5 C 330 1.4 C 81.2 C 360 1.6 C 81.7 C 390 2.0 C 81 C 420 2.5 C 80.2 C 450 3.6C 80.0 C
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Arnesha Jackson Enthalpy and Specific Heat Page of 7 8 In the figure legend (text below the graph), explain the data in your graph. Be sure to use different symbols (and make them different colors) for the two data sets and mark the maximum (hand warmer) and minimum (cold pack) temperatures with an asterisk. Explain the symbols in the figure legend. Do NOT include a trendline or connect the data points. See other general tips for making graphs in the How to Make a Graph in Excel document located in the Introductory Materials for this lab. 8. Putting it all together! Consider the temperature changes experienced by the cold pack and the hand warmer and answer the questions below. Figure 1 represents the data collected for the Delta Temp for every 30 seconds of a cold pack and hand warmer for 450 seconds. The lowest temperature that occurred for the cold pack at 90 seconds with a temperature of -1.8 C and is marked with an asterisk (*). The highest temp occurred for the hand warmer at 240 seconds with a ten of 82.7 C and is marked as well.
Arnesha Jackson Enthalpy and Specific Heat Page of 8 8 a. What was the overall Δ T (change in temperature) for the cold pack? Show your work and include units. Note: Remember that Δ T = T f - T i b. What was the overall Δ T (change in temperature) for the hot pack? Show your work and include units. Note: Remember that Δ T = T f - T i c. Which one worked via an endothermic process? Via an exothermic process? Explain in 2 – 3 sentences. d. Which pack had the greatest change in enthalpy? Explain, in 2 – 3 sentences, how you know using your experimental data. Δ T= 3.6 C- 22.9 C= -19.3 C. Δ T= 80.0 C - 29.5 C= 50.5 Endothermic: _____ICE PACK______________ Exothermic: _______HEAT WARMER____________ The hand warmer absorbs heat due to an exothermic reaction that happens due to chemical reaction within the materials of the hand warmer. Enthalpy change of exothermic reactions is negative since heat flowed away which means a decrease in the enthalpy change. Cold pack underwent endothermic reaction where energy was taken in from its surrounding material. The enthalpy change difference for cold is smaller than the enthalpy change calculations for heat so the greatest enthalpy change would be the heating pad.