One gram of supercooled liquid zinc at 400 °C is in a container of large heat capacity. Find the entropy change of zinc during solidification Zn(s) Cp=22.4 + 0.01005 J/molk AHm= 7388 J/mole at 420 °C Zn(I) Cp= 31.4 J/molK
Q: , Problem 14: A pot containing 620 g of water is placed on the stove and is slowly heated from 18°C…
A:
Q: Calculate the total change in entropy Δ?ΔS if ?1=2.55 kgm1=2.55 kg of water at ?1=17.5 ∘CT1=17.5 ∘C…
A: Given, m1= 2.55kg m2= 5.75kg T1= 17.5 °C T2= 67.5°C Specific Heat S= 4186 J/kg K
Q: What would be an expression for the entropy change for 2.15 moles of an ideal gas that undergoes a…
A: At constant temperature the entropy ∆S=-nRlnPfPi here, Pf= final pressure= 1 atm Pi= initial…
Q: Suppose 1.99 kg of water at 16.6°C is mixed with 1.36 kg of water at 79°C. After equilib- rium is…
A: Data Given. m1 = 1.99 kg T1 = 16.6°C = ( 273 + 16.6 ) K = 289.6 K m2 = 1.36 kg T2 = 79°C = ( 273…
Q: S= R+R ln 2 ² h² 2лmk T 3/2 B V o determine that the entropy content of one mole of argon gas at 300…
A: Solution: The entropy of one mol of argon gas at a temperature of 300K is 155 J/K. Then the entropy…
Q: On a winter day, a certain house loses 1.12 x 108 J of heat to the outside. What is the total change…
A:
Q: 20 grams of ice initially at -15◦C is placed on top of 100 grams of 350◦C solid gold in a thermally…
A:
Q: A container with 41 g of a certain liquid (c = 4171 J/kg · °C, Lv=2224 kJ/kg) at 22 °C is heated up…
A:
Q: Junyoung and Enkhtur are experimenting with an ideal gas. in their experiment, a 573-g quantity of…
A:
Q: A piston/cylinder contains air assumed to be an ideal gas at 800 kPa and 300 K. The gas undergoes…
A: a) According to first law of thermodynamics dQ = dU + PdV Since the process is isothermal dU = 0…
Q: A cylindrical copper rod of length 1.50 m and radius 2.00 cm is insulated to prevent heat loss…
A: The given values are,
Q: 20 g copper at 40 degree celsius is dropped into a constant volume calorimeter containing 100 g…
A:
Q: A 1.0 mole sample of a “monatomic” ideal gas undergoes a reversible isobaric expansion from Vi to…
A: The change in entropyΔS of the gas.given,no. of moles=1Initial volumeVi=Vifinal…
Q: À 1.20 mol of an ideal gas at 350 K undergoes a free adiabatic expansion from V1 = 12.3 L to V2=…
A:
Q: A litre of water at 20 degrees C is placed in a fridge at 5 degrees C. Calculate the change in…
A: When an object or a body loses heat, its temperature falls and the heat loss is mathematically…
Q: A 5 kg block of ice initially at 268.15 K is placed in thermal contact with a very large reservoir…
A:
Q: The temperature at the surface of the Sun is approximately 5,600 K, and the temperature at the…
A:
Q: How much energy must be transferred as heat for a reversible isothermal expansion of an ideal gas at…
A:
Q: The surface of the Sun is approximately at = 700 K, and the temperature of the Earth’s surface is…
A:
Q: Three moles of an ideal gas underrgo reversible isothermal compression at 20 degrees C. During…
A:
Q: A 68.7-g ice cube is initially at 0.0°C. (a) Find the change in entropy of the cube after it melts…
A: The given data are: m=68.7 g=0.0687 kgLf=3.33×105 J/kgT=0° C=273 K Here, m denotes the mass, T…
Q: QUE STION FOUR: A light bulb operates at a temperature of 4,300 K and has an emissivity of 0.800 and…
A: Given, Temperature, T=4300K Area, A=5.5×10-6m2 Mass, m=2×10-3kg Initial Temperature, Ti=-30°C Final…
Q: . One mole of an ideal gas expands isothermally at T = 20°C from 1.1 m³ to 2.1 m³. The gas constant…
A:
Q: Heating 4.5 mol O₂ at a constant external pressure results in the temperature changing from 264K to…
A: We are given the value of Cp as function of time. We know that change in entropy for small change in…
Q: 30 grams of ice at -5 Celsius melt while being left at room temperature of 28 Celsius. After…
A: Given data The mass of the ice is m = 30 grams The initial temperature of ice is Ti-ice = -5⁰C = 268…
Q: A glass of 450.0 g water at 20°C is chilled using a 20.0 g icecube at 0°C. Assume that the mixing…
A:
Q: A 1.20 mol of an ideal gas at 350 K undergoes a free adiabatic expansion from V1 = 12.3 L to V2 =…
A:
Q: Suppose you have a 36.4-g sample of a protein for which denaturation required an input of 2.20 J at…
A:
Q: Find the change in entropy of a monoatomic Van der Waals gas using the equation of state nRT an² an²…
A: The objective of the question is to find the change in entropy (ΔS) of a monoatomic Van der Waals…
Q: An 8.0 g ice cube at -10˚C is put into a Thermos flask containing 110 cm3 of water at 15˚C. By how…
A: Given that:Volume of water, Vw=110 cm3Mass of water, mw=110 cm3 × 1 g/ cm3=110g= 0.11 kgMass of ice,…
Step by step
Solved in 3 steps with 1 images
- One mole of a monoatomic ideal gas initially has a volume of 24.4L at latm and 298K undergoes a cyclic process as shown in Figure 6. Process 2→3 is an adiabatic expansion and process 3→1 is an isothermal compression. Given the V3 = 138L and T2 = 596K 2 1 3 Volume Figure 6 Calculate (a) the internal energy change in the process 1>2 (b) the entropy change in the process 1→2 (c) the heat released in the process 3->1 (d) the internal energy for one complete cycle 1→2→3→1. (e) the total work done for the entire process. PressureFor ice melting into water at 0 °C, the latent heat of fusion is Lp = 3.34 x 105 J/kg. For water boiling at 100 °C, the latent heat of vaporization is Ly = 2.26 x 105 J/kg. Which process involves a greater change in entropy? O melting 1.00 kg of ice at 0 °C O vaporizing 1.00 kg of liquid water at 100 °C the entropy change is the same for bothWhat is the entropy change for 3.20 mol of an ideal monatomic gas undergoing a reversible increase in temperature from 380 K to 425 K at constant volume?
- What is the minimum change of entropy that occurs in 0.200 kg of ice at 273 K when 7.40 x 104 J of heat is added so that it melts to water? O 2.71 x 10² J/K 3.69 x 10-³ J/K 7.4 x 104 J/K 3.11 x 10² J/K O 3.22 x 10-³ J/KFind the change in entropy of a monoatomic Van der Waals gas using the equation of state nRT an² an² P = and U = nRT V-nb V2 V (Hint: Start from AU = TAS-pdV)Find the change in entropy that results when a 3.2-kg block of ice melts slowly (reversibly) at 1 °C. Assume that Latent frequency is 3.35 x 10^5 J/Kg.
- 30 grams of ice at -5 Celsius melt while being left at room temperature of 28 Celsius. After melting its temperature rises to 28 Celsius. Determine the total change of entropy of the ice. [Cice = .5 cal/g.C, Lf = 79.6 cal/gram] 1 calorie = 4.184 J Options: A. 49.97 J/K B. 34.68 J/K C. 53.11 J/K D. 38.27 J/K E. 44.22 J/KWhat is the entropy change when 0.50 moles of CHCl₃ vaporizes at 61.0 °C? [∆H(vap) = 29.6 kJ/mol at 61.0 °C] ____J/K