### Chapter 8, second assignment

Answers to selected problems for Chapter 8 are found here.

 Exercise 19. If we warm a volume of air, it expands. Does it then follow that, if we expand a volume of air, it warms? Explain. No. By warming a volume of air, we are putting energy in; the energy goes into expansion, and therefore expansion requires energy. If we expand a volume of air without putting energy in, we are actually withdrawing energy from the air and it will cool down! Exercise 20. Machines used for making snow at ski areas blow a mixture of compressed air and water through a nozzle. The temperature of the mixture may initially be well above the freezing temperature of water, yet crystals of snow are formed as the mixture is ejected from the nozzle. Explain how this happens. The loss of pressure on the compressed air causes it to expand rapidly; hence it cools rapidly. The rapid cooling is what produces the snow crystals. A CO2 fire extinguisher does the same thing: carbon dioxide at room temperature, but under a lot of pressure, expands rapidly out the nozzle and cools enough that you actually form dry ice snow (frozen CO2)! Exercise 35. Why will wrapping a bottle in a wet cloth at a picnic often produce a cooler bottle than placing the bottle in a bucket of ice water? Evaporation rapidly withdraws heat from the "bottle plus cloth" ensemble. On the other hand, a bucket of ice water will cool the bottle but only by conduction; neither water nor ice is a good conductor of heat! Exercise 39. Place a jar of water on a small stand within a saucepan of water so that the bottom of the jar is held above the bottom of the pan. When the pan is put on a stove, the water in the pan will boil, but not the water in the jar. Why? The water in the pan is warmed to (and not over) 100°C by the much hotter stove, in spite of losses to the air above the water. However, the water in the jar is warmed only by the hot water in the pan. It cannot get hotter than 100°, and because it is continually losing heat to the air above it, it cannot even get to 100°! This principle has been known (as a practical technique for gentle heating) for many centuries; it's sometimes called a bain-marie after the fourth-century alchemist Mary the Jewess. In cooking, this sort of thing is called a double boiler. Exercise 40. Water will boil spontaneously in a vacuum--on the moon, for example. Could you cook an egg in this boiling water? Explain. No. Cooking depends on temperature; water boiling in a background is boiling at or below "room temperature" and you'd never get your egg done! Exercise 44. In the power plant of a nuclear submarine, the temperature of the water in the reactor is above 100°C. How is this possible? Pressurizing the water keeps it from flashing to steam; with pressure, the boiling point will rise. Exercise 46. Why is it that, in cold winters, a tub of water placed in a farmer's canning cellar helps prevent canned food from freezing? The freezing water gives off heat that keeps the cellar above the (rather lower) freezing point of the canned food. Exercise 48. Why does a dog pant when it is hot? The dog is using evaporative cooling in its lungs. Panting moves air rapidly in and out of the moist breathing passages inside the dog and keeps its core temperature down. Problem 2. Calculate the height from which a block of ice at 0°C must be dropped for it to completely melt upon impact. Assume that there is no air resistance and that all the energy goes into melting the ice. [Hint: equate the joules of gravitational potential energy to the product of the mass of ice and its heat of fusion, 335,000 J/kg. Do you see why the answer doesn't depend on mass?] The gravitational potential energy is equal to mgh, where m is mass, g is the acceleration of gravity (9.8 m/s2) and h is the height--the thing we want to find. The heat required to melt the ice is equal to Lfm, where Lf is the heat of fusion, and m is again the mass. Setting these equal to each other and allowing the masses on each side to cancel, gh = Lf 9.8 m/s2 × h = 335,000 J/kg But 1 J = 1 kg m2/s2, hence 1 J/kg = 1 m2/s2. Therfore, 9.8 m/s2 × h = 335,000 m2/s2 h = 335,000 m2/s2 ÷ 9.8 m/s2 h = 34,200 meters (!!) 34.2 km = 20.5 miles. Problem 3. The specific heat capacity of ice is about 0.5 cal/g/°C. Supposing that it remains at that value all the way to absolute zero (-273°C), calculate the number of calories it would take to change a 1-gram ice cube at absolute zero to 1 gram of boiling water. How does the number of calories compare with the number of calories required to change the same gram of 100°C water to 100°C steam? The key numbers are found in the box on p. 197: heat capacity of water Cw = 1 cal/g/°C heat capacity of ice Ci = 0.5 cal/g/°C heat of fusion of water Lf = 334 J/gheat of vaporization of water Lv = 2256 J/g You also need to know that 1 cal = 4.814 J. There are three parts to our first calculation: How much heat is required to take 1 g of ice from absolute zero to 0°C?How much heat is required to melt 1 g of ice?How much heat is required to take 1 g of water from 0°C to 100°C? The amount of heat required to change the temperature of 1 g of ice by 273°C is given by Q1 = CimΔT. That works out to be 0.5 cal/g/°C × 1 g × 273°C = 136.5 cal. The amount of heat required to melt 1 g of ice at 0°C is given by Q2 = Lfm, which works out to be 334 J or 79.83 cal. The amount of heat required to change the temperature of 1 g of water by 100°C is given by Q3 = CwmΔT. That works out to be 1 cal/g/°C × 1 g × 100°C = 100 cal. The total heat required to bring 1 g of ice/water from absolute zero to 100°C is the sum Q1 + Q2 + Q3 = 316.3 cal. The second calculation is simply to use the heat of vaporization of water to determine the energy needed to change 1 g of water into 1 g of steam: Qv = Lvm. For 1 g of water, we therefore need 2256 J of heat, which is to say, 539.2 cal. This is more than one-and-a-half times the heat required to bring the same mass of water from absolute zero to the boiling point, and explains why we can boil water for quite some time before it boils away.