MidtermExam_260_Spring_2024_solutions

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© Mark Schlossman – do not copy or post without permission 1 PHYSICS 260 NAME SOLUTIONS Midterm Exam September 29, 2023 All work must be shown to receive full credit. Work must be legible and comprehensible. Answers, including units , must be clearly indicated. Equation Sheet is on the last page. PROBLEM POINTS SCORE 1 16 2 20 3 26 4 20 5 18 TOTAL 100

© Mark Schlossman – do not copy or post without permission 2 1. (16 points) Multiple Choice – Circle the correct answers (for example circle the label “i”). More than one choice may be correct for each problem. Note that points will be taken off for wrong answers. (a) (4 points) Write a mathematical expression for the First Law of Thermodynamics. ___ D U = Q – W by __________________________________ (b) (4 points) Equipartition i. refers to the distribution of energy in a system at thermal equilibrium. ii. refers to energy units, independent of the temperature, to a system’s degrees of freedom. iii. refers to assigning energy units to translational motion, then if any energy remains, assigning energy units to rotational and vibrational motion. iv. can be used to relate the internal energy U of a system to its temperature. v. says that two systems placed into thermal contact reach equilibrium by making equal changes in temperature. (c) (4 points) Heat capacity i. is the amount of heat in a system. ii. is proportional to the number of degrees of freedom in a system. iii. is used to determine the amount of heat required to change the temperature of a system. iv. has units of joules/(area second). v. is relevant only for processes at constant volume. (d) (4 points) For a monatomic ideal gas in thermal equilibrium at high temperature (> 1000 K) , i. degrees of freedom include translational, rotational, and vibrational motions. ii. U = 3/2 NkT . iii. U = 7/2 NkT . iv. U = 7/2 pV .

© Mark Schlossman – do not copy or post without permission 3 2. (20 points) A copper beaker weighing 150 g contains 10 g of water at 15 °C. 50 g of water at 90°C are poured into the beaker and the final temperature is 65.7 °C. What is the specific heat capacity of copper? Assume that heat is not exchanged with the surroundings. Hint: Don’t forget that the copper beaker will change temperature along with the water and must be considered in the heat calculations. Solve this by considering an expression for the changes in internal energy D U of the 3 components (beaker, 10 g water, and 50 g water). Remember to include units in your answers and show work to earn full or partial credit. Answer: We assume that there are no thermal exchanges with the surroundings and that no work is done. This calculation can also be done without converting to K. Δ U = 0 = Δ U Cu + Δ U 10g water + Δ U 50g water c Cu (0.150 kg) ( T F − 288.15) + c w (0.01 kg) ( T F − 288.15) + c w (0.05 kg) ( T F − 363.15) = 0 where all the temperatures are in Kelvin. Using 338.85 K for the final temperature and 4.18 kJ K -1 kg -1 for the specific heat capacity of water yields c Cu = − 4.18(0.01 kg) (338.85 − 288.15) − 4.18(0.05 kg) (338.85 − 363.15) (0.150 kg) (338.85 − 288.15) = 0.389 kJ K -1 kg -1

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