1. The specific heat of silver is 0.05 cal/(g C) and the specific heat of water is 1 cal/(g C)

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1 IDS 10 Winter 009 Additional practice questions for Exam 1 1. The specific heat of silver is 0.05 cal/(g C) and the specific heat of water is 1 cal/(g C) If a certain amount of heat is added to a sample of water, the water temperature raises 5 degrees Celsius. If an equal amount of heat is added to an equal mass of silver, what would be the change in temperature of the silver? The ratio of the specific heat of water to silver is 0:1. This means that if water increases 1 degree, the silver will increase 0 degree. Therefore, if the water increases 5 degrees, the silver will increase 100 degrees C.. One of the remarkable scientific discoveries in the past twenty-five years is the discovery of creatures that live on the floor of the ocean about 00 miles west of Washington under 00 meters of ocean water. (1 kg =. lbs and 9.7 inches = 1 meter) a. The density of seawater is about 107 kg/m. What is the pressure of the ocean on these creatures in psi? Show your work. The pressure exerted by the seawater is equal to the density times the height: kg kg a. 107 x00m,6,100 m m But these are not the correct units for pressure. Converting to units of psi. kg.lbs 1m,6,100 x x m 1kg ft 1 ft x 144in lbs 54 in Note: there are other possible ways to approach this problem. This is just one example. b. In class, we did some calculations of the pressure of the atmosphere at sea level. Since these creatures are under the ocean and the atmosphere, what is the total pressure they experience? The total pressure is the sum of the pressure due to the ocean water (from part a) and the pressure due to the atmosphere. So P = = 68.7 psi Note that the atmospheric pressure makes a very small contribution to the total pressure at these depths! c. Even if we (humans) have a source of air, we could not survive the pressure at this depth of the ocean without a submarine. How do these organisms survive this pressure?

2 We survive at pressures of 14.7 psi because the pressure inside our body is roughly equal to the pressure outside. The same must be true for organisms on the ocean floor. If the pressure due to the water was greater than their internal pressure, they would be crushed.. The following statement was found on a web site. Review the statement and decide if the statement is correct as written. (Note: there may or may not be errors.) Show your work/explain your reasoning. How much pressure are you under? Earth's atmosphere is pressing against each square inch of you with a force of 1 kilogram per square centimeter (14.7 pounds per square inch). The force on 1,000 square centimeters (a little larger than a square foot) is about a ton! Checking the math we find.. 1kg.lbs 1000cm x x 000lbs cm kg Since 000 lbs = 1 ton, the calculation is roughly correct. The force (i.e. mass in lbs) on 1000 cm is about a ton. However, note that the authors use units of kg/cm to describe a force. Kilograms are a unit of mass not force. And even if kg were a unit for force, then units of kg/cm would be units for pressure. 4. Imagine two tall glass tubes filled with water. Both tubes are 50 inches high, but one tube has a diameter of in, while the other has a diameter of 4 in. Compare the pressure exerted by the water in each of these tubes. Clearly explain your reasoning. As we learned, the pressure exerted by a solid or liquid depends only on the density and the height. Since these two tubes are the same height and both contain water, they exert the same pressure regardless of the diameter. The fatter tube exerts a greater force, but that force is spread out over a larger area. 5. An architect is designing a building and needs your help. She is designing a structure that will have numerous 4-inch diameter columns that are 0 feet high. She is concerned about the pressure each column will exert on the marble floor. She is thinking about making the columns out of three different materials, wood, granite, and water (in our calculations we will ignore the pressure created by the container which will hold the water). a. Determine the pressure of each type of material in pounds per square inch. (The density of wood is about 0.5 grams/cm, granite is.6 grams/cm, and the water is 1.0 gram/cm and the formula for the volume of a cylinder is r h). Pressure is Force per unit area and can be expressed in units of pounds per square inch. To find the pressure we need both the mass of the material, and its cross sectional area. The mass can be found from the density and the volume ( mass = density x volume)

3 The volume of a cylinder is A x h. We have to be very careful about the units. Since the density is in units of g/cm, and we want the pressure in units of pounds/in, we should start by converting the density to units of lbs/in. (See problem b). For water (density = 1 g/cm ) g 16.9cm 1kg 1lb 1.0 x x x cm 1in 1000g 0.45kg For wood (density = 0.5 g/cm ) g 16.9cm 1kg 1lb 0.5 x x x cm 1in 1000g 0.45kg lb 0.06 in lb in (here s a simpler way wood is half as dense as water, so the density of wood = (1/)x(0.06 lbs/in ) = lbs/in For granite (density =.6 g/cm ) g 16.9cm 1kg 1lb lb.6 x x x cm 1in 1000g 0.45kg in (again. a simpler way..granite is.6 times as dense as water, so the density of granite = (.6)x(0.06 lbs/in ) = lbs/in We will use the density to find the mass of each column, but first we need the volume. To make the units work out, we will need the volume in units of in. Recall that for a cylinder, volume = A x h. The radius of each column is 1 in The cross sectional area of each column is r = x (1) = 45 in The height is 0 feet x (1 in /1 foot) = 40 in Thus, the volume = V = A x h = in Now for the mass. Mass = density x volume. For water, the mass is 0.06 lbs/in x in = 905 lbs For wood, the mass is lbs/in x in = 1950 lbs For granite, the mass is lbs/in x in = 1000 lbs Finally, the Pressure is F/A For water, P = 900 lbs/ 45 in = 8.6 lbs/in or 8.6 psi For wood, P = 1950 lbs/ 45 in = 4. lbs/in or 4. psi For water, P = 1000 lbs/ 45 in = lbs/in or psi That was complicated! Here is an easier way. Did you notice that we used the cross sectional area twice? Watch carefully and see how it cancels out. Force P, right? Area And.. Force ( mass) Volumex Density But. Volume = Area x height So.Force = Area x height x density which means that Look! The Area cancels out. Which means that. Pressure = height x density P Area x heightx density Area

4 Let s try it out. For wood..p = (40 in) x (0.018 lbs/in ) = 4. psi b. After you calculated part a above, she wants to know what the pressure (in psi) would be if the columns are 1 inches in diameter instead of 4 inches. See the answer to problem A slab of bedrock lies underneath 40 meters of soil and 100 meters of ice. The density of the soil is. grams per cubic centimeter. The density of ice is 0.9 grams per cubic centimeter. The situation is shown in the diagram below. ice rock soil a) Find the pressure exerted on the slab of rock. (It is probably easiest to answer in units of bars where one bar is one kg per square centimeter which is approximately atmospheric pressure.) The ice and the soil both weigh down on the rock below. We need to consider the contribution from each and then add them together. We could look at a one square centimeter portion of the rock and calculate how much weight is pushing down on it, or we could use the pressure = density height idea. I ll do the former. First some conversions: 100 meters is 10,000 cm and 40 meters is 4,000 cm. For the ice: above a one square centimeter patch of rock we have (1 cm )(10,000 cm)(0.9 g/cm ) = 9,000 g = 9 kg of ice. For the soil: above a one square centimeter patch of rock we have (1 cm )(4,000 cm)(. g/cm ) = 9,00 g = 9. kg of soil. The pressure exerted on the rock by the solid stuff above is the sum of the two contributions above: (9 kg of ice over each cm ) + (9. kg of soil over each cm ) = 18. kg The pressure is 18. kg/cm, which is the same as 18. bar (or about 18 atmospheres). b) Did you use atmospheric pressure to find the answer to part a? Should you? Why or why not? Explain your reasoning. Well, gosh. No I didn t. If we add atmospheric pressure, that makes the answer 19. bar. That s a big enough change that we probably should have taken it into account.

5 c) How much of the pressure that you found in part a was due to the soil? How much was due to the ice? From part a we see that 9 bar was from the ice and 9. bar was from the soil. d) Imagine that the situation were somewhat different. Imagine that instead of a layer of pure soil 10 meters thick below a layer of pure ice 0 meters thick, the actual situation was that there was the same amount of ice and the same amount of soil but they were all mixed together. Would there be any difference in the total pressure exerted on the slab of rock? Would there be any difference in the fraction of that pressure that was exerted by ice? by soil? Explain your reasoning. The total weight of stuff above any square centimeter of rock would be the same, so the pressure exerted on the rock would be the same. Since the total weight of ice is the same and the total weight of soil is the same, the ice still exerts a pressure of 9 bar and the soil still exerts a pressure of 9. bar (and the atmosphere exerts a pressure of 1 bar). The total pressure is still 19. bar. e) Now think of the atmosphere. The atmospheric pressure on your body right now is about 14.7 pounds per square inch. The atmosphere around you is about 79% nitrogen, 0% oxygen, and 1% water vapor (this last number is a guess the other two don t change much). How much of the pressure that is exerted on your body right now is due to nitrogen? The total pressure is 14.7 psi. Nitrogen accounts for about 79% of that. 79% of 14.7 psi is 11.6 psi. Don t memorize this number, but this means that under these circumstances the partial pressure of nitrogen in the air is 11.6 psi. f) How much of the pressure that is exerted on your body right now is due to oxygen? Oxygen accounts for about 0% of 14.7 psi. 0% of 14.7 psi is.9 psi. This means that under these circumstances the partial pressure of nitrogen in the air would be about.9 psi. g) How much of the pressure that is exerted on your body right now is due to water vapor? 1% of 14.7 psi is 1.5 psi. The partial pressures of nitrogen and oxygen are always about the same fraction of the atmospheric pressure. The partial pressure of water vapor varies a lot. It has a major influence on weather and climate everywhere.

6 7. Two people calibrated their own thermometers (and made up their own temperature scales) in different locations. They both used the temperature of the sea as their temperature of 0. They both used their body temperatures as temperatures of 100. Hal created "degrees Hal" ( H) in Hawaii and Will created "degrees Will" ( W) in Washington. As a result of the differences in water temperature, 0 H = 17 C and 0 W = 7 C, but 100 H = 7 C and 100 W = 7 C. a) What would a temperature of 10 H be in degrees Celsius? The range of deg H is 100 deg, while in deg C the range is 0 deg. Therefore, for every deg C, there are 5 deg H. A temperature of 10 deg H would be 17 deg C plus deg= 19 deg C b) What would a temperature of 10 H be in degrees Will? From part a) we found that 10 deg H is the same as 19 deg C. In the W scale there are 100 degrees for the range of 0 degrees in the C scale. Therefore every deg C =. deg W. 19 deg C is 1 deg C above 7 deg C. So, 1 deg C times. = 9.6 or 40 deg W. c) Which is bigger: one degree Hal or one degree Will? If 5 deg H = 1 deg C and. deg W = 1 deg C, then 5 deg H =. deg W. so, the degrees W must represent a large increment of temperature increase per degree.. 8. If we want to make jewelry with gold, we would combine the gold with copper or gold (4k is pure gold and lower karat values indicates a higher percentage of either copper or silver mixed with the gold). In this question we are interested in pure gold. A 150 g sample of pure gold initially at 5 o C is heated in a furnace that supplies energy at the rate of 50 cal per minute. The melting temperature of gold is 106 degrees C The heat of fusion for gold is 16.1 cal per gram The specific heat of gold is 0.0 cal/g deg C a. Sketch a graph of temperature vs. time, showing how the temperature of the gold changes over a time period of 5 minutes. Explain your reasoning.

7 Temperature of the gold (degrees C) Change in temperature of gold over 5 minutes Series Time (minutes) b. After 5 minutes of gaining heat, how much (if any) of the gold will be melted. Show your work and explain your reasoning. To get the gold to the melting temperature: H = Mass X change in temperature X specific heat H= 150 g X 108 deg C X 0.0 cal/g deg C H= 4671 calories The oven supplies 50 calories per minute, therefore the melting temperature will be achieved in 4671 calories/ 50 calories/min = 18.7 minutes. This means that we have 6. minutes of heat remaining to melt the gold. The heat of fusion for gold is 16.1 cal/g. If the oven will supply 6. minutes of heat and it is 50 calories per minute, then there will be 1575 calories added to the gold. To determine the amount of gold that will melt: 1575 calories/ 16.1 calories/ g = 97.8 grams of gold will melt. Notice that the temperature of the gold does not change during the melting process.

8 c. Imagine that we have formed a gold ring and permitted it to solidify and cool to a temperature of 800 degrees C. If we put this 800 degree C gold ring in 100 grams of 5 deg C water, will any of the water vaporize? If so, how much water will vaporize. If not, explain why no water vapor would form. Let s determine the heat that would be supplied by the gold: (We will assume that if any water is vaporized the temperature of the gold can not be below 100 deg C.) Heat = 150 g X (800 deg C-100 deg C) X 0.0 cal/g deg C Heat= 150 calories To get to the boiling point of water: Heat = 100 g (100 deg C- 5 deg C) = 7500 cal So, none of the water would vaporize. It requires more heat to change the temperature of the water. 9a) In Finish society, saunas are a place for improving one s health and a social gathering place. Commonly saunas are small rooms, lined in wood with a stove to heat the room. It is common to find a container of rocks on top of the stove. The following is a quote from Wikipedia: Under many circumstances, temperatures approaching and exceeding 100 C (1 F) would be completely intolerable. Saunas overcome this problem by controlling the humidity. The hottest Finnish saunas have very low humidity levels, which allows air temperatures that could boil water to be tolerated and even enjoyed for long periods of time. Other types of sauna, such as the hammam, where the humidity approaches 100%, will be set to a much lower temperature of around 40 C (104 F) to compensate. The "wet heat" would cause scalding if the temperature were set much higher. ( Explain why a sauna with high humidity at high temperatures (for example, 80 deg C) could create scalding, while a sauna with low humidity can be enjoyable at this same temperature. There are a couple of reasons for this concern. 1) When we get hot, we sweat. The moisture on our skin evaporates and takes heat energy from our bodies with it. Higher humidity means that the moisture on a person s skin is less likely to evaporate. A person s body temperature increases to a point that heat stroke is possible.

9 Another reason is that when the water in the air of the humid sauna condenses on a person s skin, 540 cal per gram is liberated and some of that thermal energy is absorbed by the skin therefore the scalding. 9b) Sven decides to add some water to the rocks on the top of the stove in his sauna (this is a common practice in Finland). The rocks on the stove are slightly hotter than the room (assume that the rocks are 110 deg C). If there is 10 Kg of rock and 100 ml of 80 deg C water, will all of Sven s water vaporize when he throws the water on the rocks? (Assume that the specific heat of the rocks is 0. cal/g deg C and the heat of vaporization for water is 540 cal/g). Like many of these types of problems, it is good to divide them into parts: Determine the amount of heat from the hot rocks--- (we will assume that the rocks go down to 100 deg C. If the rocks when below 100 deg C, there would be not vaporization of the water). 10 Kg = 10,000 g Heat = (10000 g) (10 deg C) (0. cal/g deg C) = 0,000 cal How much heat is required to change the temperature of the water from 80 deg C to 100 deg C? 100 ml = 100 g for water Heat = (100g) (0 deg C) (1 cal/g deg C) = 000 cal The last step is to determine how heat is necessary to vaporize the water: 100 g (540 cal/g) = 54,000 cal So, it takes 54,000 cal to vaporize all the water. It is clear that not all of the water will vaporize because only 18,000 cal would be available. (Although not a part of the question it would vaporize 1/ of the water about g. The remaining water would eventually vaporize as the stove heats the rocks again.) 10. When nitrogen in the air is condensed into a liquid we call it (big surprise!) liquid nitrogen. Liquid nitrogen is very cold and is used by doctors to treat some skins conditions. We have invented a new temperature scale where the zero point was chosen to be absolute zero and the temperature of ten degrees was chosen to be the boiling point of water. Since the temperature scale was invented at Green River, we can refer to it as degrees Green or G. The table below compares the Green temperature scale with the Kelvin and Celsius scales.

10 Degrees Green Degrees Celsius Degrees Kelvin Absolute zero 0 G -7 C 0 K Water boils (at 1 Atm) 10 G 100 C 7 K a) The temperature on Pluto (the object formerly known as a planet, not the dog) is 7 K. What is that in degrees Green? Plotting the Green scale along the y-axis and the Kelvin scale along the x-axis, we get a slope of 10/7 = 0.07 Y = mx + b Y = 0.07 x + b To find b: 0 = 0.07 (0) + b, so b = 0 Y = (0.07) (7 deg K) = 0.99 deg G or about 1 deg G b) Aluminum melts at a temperature of 660 C. What is the melting temperature of aluminum in degrees Green? Using y-y1/ x-x1 we get the same slope as the previous problem Solving for b we find that b = 7. Y = (0.07 (660) + 7. Y= about 5 deg G c) The forecast for tomorrow calls for a high temperature of 78 F (in Honolulu). As a reminder, water freezes at F and water boils at 1 F. So what is 50 F in degrees Green? We need the freezing temperature of water in deg G Y= (0.07) = 7. deg G So now we know that deg F = 7. deg G and 1 deg F = 10 deg G Slope will be.7/180 = Y= (0.015) x + B

11 10 = (0.015) (1) + B B = = 6.8 now we find the temperature of 50 deg F in deg G Y= (0.015) (50 deg F) Y= 7.55 deg G 11. Of the following equations, a through e, some are correct and some are incorrect. Some of them deal with things that should be familiar and one of them deals with things that are probably strange to you. For each equation indicate whether it is correct or incorrect and if it is incorrect write a corrected version in the space provided. You may answer these questions from your own knowledge but you may also use the information on units in the following table! You probably need to use the table for quantities that are unfamiliar. Quantity Heat of vaporization Specific heat Units calories per gram calories gram( C) Density grams per cm Pressure Insulation (as in a warm jacket) force area seconds calories For each equation below, indicate whether it is correct or incorrect. If it is incorrect, give a corrected equation in the space provided. d) Heat ( heat of vaporizati on)( mass)( T) Choose one: correct incorrect If incorrect, write a corrected equation involving heat and heat of vaporization here: Heat = heat of vaporization (mass) C e) Heat ( specific heat)( mass)( T) Choose one: correct incorrect

12 If incorrect, write a corrected equation involving heat and specific heat here: f) Volume density mass Choose one: correct incorrect If incorrect, write a corrected equation involving volume and density here: Volume = mass/density g) Force pressure area Choose one: correct incorrect If incorrect, write a corrected equation involving pressure and area here: h) Insulation heat lost temperature change time taken Choose one: correct incorrect Since we haven t learned about insulation, you do not need to provide a corrected equation here. Simply explain your reasoning in deciding whether this is probably correct or incorrect. Reasoning: