<p>Thermodynamics Practice Problem Set - With Solutions in green</p><p>Question 1 (1 point)</p><p>What are the Zeroeth, First, and Second Laws of Thermodynamics?</p><p> Zeroth law (Thermal Equilibrium): A=B, A=C Then also B=C</p><p>First law (law of conservation of energy): Heat = Energy, </p><p>Second law: Total entropy of the system increases in any irreversible process</p><p>Not needed for the question but included in the solutions for study purposes: Third law: As the temperature approaches zero, the entropy approaches zero Question 2 (3 points)</p><p>On the planet Zardax, the temperature scale has the boiling point of water (our 212 F) at zero Zardaxian degrees and the freezing point of water (our 32 F) at negative 120 Zardaxian degrees. What is the value of absolute zero in Zardaxian degrees?</p><p>-273 0 100 ºZ ? -120 0</p><p> It is easier to work from the Celsius scale than the Fahrenheit scale; the above table shows the values of absolute zero, the freezing point of water, and the boiling point of water on the Celsius and Zardaxian scales. The relationship between the two scales is linear; i.e. of the form y = mx +b</p><p>Z = mC + b</p><p>-120= m(0º) +b, b = -120</p><p>Z = m(100º)-120 = 0</p><p>100m = 120</p><p> m= = 1.2</p><p>Z = 1.2C -120</p><p>Z = 1.2(-273) -120 = -448 ºZ</p><p>Question 3 (3 points)</p><p>How much energy is required to heat a 12 kg block of ice from -75 degrees Celsius until it becomes steam at 180 degrees Celsius? What is the change in entropy that takes place during the melting phase? What is the change in entropy that takes place during the boiling phase? -75 ------> 0º ------> 0º ------>100º ------> 100º ------>180º</p><p> ① Q ICE = mcΔT</p><p>= (12kg) (2090 J/kg◦C)(75◦C) = 1881KJ</p><p>② Qwater = (12kg)(4186 J/kg◦C)(100◦C) = 5023KJ</p><p>③ Qsteam = (12kg)(2010 J/kg◦C)(80◦C) = 1930KJ</p><p>Q = mL</p><p>Qmelting = (12kg)(3.35x105 J/kg) = 3996KJ</p><p>Qboiling = (12kg)(2.26x106 J/kg) = 27120KJ</p><p>The total of the five heats is 39950 KJ or 40000 KJ when significant figures are taken into account</p><p><Melting></p><p>ΔS = = = 14800J/K</p><p><Boiling></p><p>ΔS = = = 72700J/K</p><p>Question 4 (3 points)</p><p>What is the average (root mean square) speed of Helium atoms contained in a vessel with a volume of 2 cubic meters, in which the gas has a temperature of 50 degrees Celsius?</p><p> T =50 </p><p>VRms = = = 1415 m/s</p><p>Note: for O2 at 0º Celsius the rms speed of the gas is 460 m/s. The higher speed for the helium in this question comes mostly from the relatively low mass of the Helium atom. </p>