phys 4b Assignment 1 - Chapter 17.pdf

phys 4b Assignment 1 - Chapter 17.pdf - Assignment 1 ­...

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Unformatted text preview: 8/28/2015 Assignment 1 ­ Chapter 17 Assignment 1 ­ Chapter 17 Due: 1:00pm on Monday, August 31, 2015 You will receive no credit for items you complete after the assignment is due. Grading Policy Item 1 You work in a materials testing lab and your boss tells you to increase the temperature of a sample by 43.5 ∘ C . The only thermometer you can find at your workbench reads in degrees Fahrenheit. Part A If the initial temperature of the sample is 68.5 ∘ F , what is its temperature in degrees Fahrenheit when the desired temperature increase has been achieved? Express your answer in degrees Fahrenheit to three significant figures. Hint 1. How to approach the problem Recall that the equation T F = (9/5)T C + 32 used to convert one absolute temperature to another is based on the freezing and boiling points of water. The multiplicative factor of 9/5 represents the difference in a change of 1∘ F compared to a change of 1∘ C , whereas the additive factor of 32 represents the difference in the freezing point of water on the two scales, which is 0∘ C or 32 ∘ F. Hint 2. Calculate the temperature change in Fahrenheit Calculate the temperature change ΔT F of the sample in degrees Fahrenheit instead of degrees Celsius. Express your answer in degrees Fahrenheit to three significant figures. ANSWER: ΔT F = 78.3 ∘ F Incorrect; Try Again; 3 attempts remaining ANSWER: TF = 147 ∘ F 1/15 8/28/2015 Assignment 1 ­ Chapter 17 Correct Item 2 The supersonic aircraft Concorde has a length of 62.7 m when sitting on the ground on a typical day when the temperature is 14.0 ∘ C . The Concorde is primarily made of aluminum. In flight at twice the speed of sound, friction with the air warms the Concorde's skin and causes the aircraft to lengthen by 26.0 cm . (The passenger cabin is on rollers, so the airplane expands around the passenger cabin.) Take the coefficient of linear expansion for aluminum to be α = 2.40×10−5 /∘ C . Part A What is the temperature T of the Concorde's skin in flight? Hint 1. How to approach the problem Use the equation for the linear thermal expansion of a material to calculate the temperature change of the Concorde's skin when in flight, then use this to find its final temperature. Hint 2. Calculate the temperature change Calculate the temperature change ΔT of the Concorde's skin when in flight. Hint 1. Equation for linear thermal expansion The equation for linear thermal expansion of an object is ΔL = αL0 ΔT , where 26.0 cm is the change in the length of the object, 2.40×10−5 /∘ C is the coefficient of linear expansion, which depends on the type of material being heated or cooled, and ΔT is the change in the temperature of the object. ANSWER: ΔT = 173 ∘ C ANSWER: 2/15 8/28/2015 Assignment 1 ­ Chapter 17 T = 187 ∘ C Correct Item 3 The tallest building in the world, according to some architectural standards, is the Taipei 101 in Taiwan, at a height of 1671 feet. Assume that this height was measured on a cool spring day when the temperature was 13.5 ∘ C . You could use the building as a sort of giant thermometer on a hot summer day by carefully measuring its height. Suppose you do this and discover that the Taipei 101 is 0.485 foot taller than its official height. Part A What is the temperature, assuming that the building is in thermal equilibrium with the air and that its entire frame is made of steel? ANSWER: T = 37.7 ∘ C Correct Item 4 A crate of fruit with a mass of 39.5 kg and a specific heat capacity of 3700 J/(kg ⋅ K) slides 8.00 m down a ramp inclined at an angle of 39.4 degrees below the horizontal. Part A If the crate was at rest at the top of the incline and has a speed of 2.90 m/s at the bottom, how much work Wf was done on the crate by friction? Use 9.81 m/s2 for the acceleration due to gravity and express your answer in joules. Hint 1. How to approach the problem If no friction were acting, then the kinetic energy of the crate at the bottom of the incline would equal the difference in gravitational potential energy of the crate between its initial and final K +U +W = K +U 3/15 8/28/2015 Assignment 1 ­ Chapter 17 positions. The nonconservative (nc) frictional force is responsible for the difference. To find the work done by friction, apply energy conservation: K i + U i + Wnc = K f + U f . Hint 2. Find the initial and final kinetic energies What is the kinetic energy of the crate before it starts to slide (K i ) and after it reaches the bottom of the ramp (K f )? Express your answer in joules as two terms separated by commas. ANSWER: , Ki Kf = 0,166 J Correct Hint 3. Find the difference between initial and final potential energy What is U f − Ui , the change in the potential energy of the crate from when it starts to slide to after it reaches the bottom of the ramp? Express your answer in joules. Hint 1. A helpful formula and a helpful diagram The difference in the potential energy of the crate is given by U f , where h is the height shown in the figure below. What is the value of h? − U i = −mgh Express your answer in meters. 4/15 8/28/2015 Assignment 1 ­ Chapter 17 ANSWER: h = 5.08 m Correct ANSWER: Uf − Ui = ­1970 J Incorrect; Try Again; 4 attempts remaining ANSWER: Wf = ­1800 J Correct The frictional force opposes the motion of the crate, so the work done on the crate by friction must be a negative quantity. Part B If an amount of heat equal to the magnitude of the work done by friction is absorbed by the crate of fruit and the fruit reaches a uniform final temperature, what is its temperature change ΔT ? Hint 1. Equation for temperature change The quantity of heat needed to increase the temperature of an object by a certain amount is given by Q = mcΔT , where m is the object's mass, c is its specific heat, and ΔT is the temperature change (in kelvins) of the object. In this case Q is a positive quantity since the temperature of the crate is increasing. ANSWER: 5/15 8/28/2015 Assignment 1 ­ Chapter 17 ΔT = 1.23×10−2 ∘ C Correct Of course, the assumptions of "total heat absorption" and "uniform temperature change" are not very realistic; still, this simplified model provides a useful reminder about the transformation of mechanical energy into thermal energy when nonconservative forces are present. Item 5 A 15.7 g bullet traveling horizontally at 873 m/s passes through a tank containing 13.7 kg of water and emerges with a speed of 539 m/s . Part A What is the maximum temperature increase that the water could have as a result of this event? ANSWER: ΔT = 6.45×10−2 ∘ C Correct Item 6 Before going in for an annual physical, a 70.0­kg person whose body temperature is 37.0∘ C consumes an entire 0.355­liter can of a soft drink (which is mostly water) at 12.0∘ C . Part A What will be the person's body temperature T f inal after equilibrium is attained? Ignore any heating by the person's metabolism. The specific heat capacity of a human body is 3480 J/kg ⋅ K . Hint 1. How to approach the problem Use the volume of the soda (assumed to be the same as water) to calculate its mass, then use this in an appropriate equation relating the temperature changes of the person and the soda. 6/15 8/28/2015 Assignment 1 ­ Chapter 17 Hint 2. Setting up the equation Since the temperature change is assumed to be solely from the difference in temperature between the soda and body (ignoring any heat generated by the body, lost to the surrounding air, etc.), whatever heat that is lost by the body must go into heating the soda until they reach a single uniform temperature. Because there is no heat entering or leaving the system, the total heat gained (or lost) by the system must be zero, so this will be the sum of the heat gained by the soda plus the heat lost by the body. For each object, Q = mcΔT , where Q is the heat gained (if positive) or lost (if negative) by the object, m is its mass, and c is its specific heat. The temperature change ΔT will be positive for the object being heated (in this case, the cold soda warmed by the body) and negative for the object being cooled (in this case, the warm body being cooled by the soda). Hint 3. Density and specific heat of water These values can be found in almost any physics textbook: The density of water is 1.00 g/cm 3 and the specific heat of water is 4190J/kg ⋅ K . = 1.00 kg/liter Hint 4. Calculate the mass of soda Calculate the mass m s of the soda, assuming its properties to be the same as water. Express your answer in kilograms. ANSWER: ms = 0.355 kg ANSWER: T f inal = 36.8 ∘ C Correct Part B Is the change in the person's body temperature great enough to be measured by a medical thermometer? (A high­quality medical thermometer can measure temperature changes as small as 0.1∘ C or less.) ANSWER: 7/15 8/28/2015 Assignment 1 ­ Chapter 17 yes no Correct A high­quality medical thermometer can measure temperature changes as small as 0.1∘ C or less. Even so, drinking an entire soda before a medical exam will not actually lower your temperature by all that much, since by the time your body has reached the equilibrium temperature, the heat generated by your body would more than make up for any heat lost to the soda. Item 7 Just about everyone at one time or another has been burned by hot water or steam. This problem compares the heat input to your skin from steam as opposed to hot water at the same temperature. Assume that water and steam, initially at 100∘ C , are cooled down to skin temperature, 34∘ C , when they come in contact with your skin. Assume that the steam condenses extremely fast. We will further assume a constant specific heat capacity c = 4190 J/(kg ⋅ K) for both liquid water and steam. Part A Under these conditions, which of the following statements is true? ANSWER: Steam burns the skin worse than hot water because the thermal conductivity of steam is much higher than that of liquid water. Steam burns the skin worse than hot water because the latent heat of vaporization is released as well. Hot water burns the skin worse than steam because the thermal conductivity of hot water is much higher than that of steam. Hot water and steam both burn skin about equally badly. Correct The key point is that the latent heat of vaporization has to be taken into account for the steam. Part B How much heat H 1 is transferred to the skin by 25.0 g of steam onto the skin? The latent heat of vaporization for steam is L = 2.256 × 10 6 . J/kg 8/15 8/28/2015 Assignment 1 ­ Chapter 17 Express the heat transferred, in joules, to three significant figures. Hint 1. Determine the heat transferred from steam to skin Find an expression for the heat H 1 transferred from the vapor to the skin. Express your answer in terms of c, the mass of steam m , the latent heat of vaporization L, and the temperature difference ΔT between the initial temperature of the steam and the skin temperature. ANSWER: H1 = m(cΔT + L) ANSWER: H1 = 6.33×104 J Correct Here we assumed that the skin continues to remain at 34∘ C . Actually the local temperature in the area where the steam condenses can be raised quite significantly. Part C How much heat H 2 is transferred by 25.0 g of water onto the skin? To compare this to the result in the previous part, continue to assume that the skin temperature does not change. Express the heat transferred, in joules, to three significant figures. Hint 1. Determine the heat transferred from water to skin Find an expression for the heat H 2 transferred from the hot water to the skin. Express your answer in terms of c, the mass of water m , and the temperature difference ΔT between the initial temperature of the water and the skin temperature. ANSWER: H2 = mcΔT 9/15 8/28/2015 Assignment 1 ­ Chapter 17 ANSWER: H2 = 6910 J Correct The amount of heat transferred to your skin is almost 10 times greater when you are burned by steam versus hot water. The temperature of steam can also potentially be much greater than 100 ∘ C . For these reasons, steam burns are often far more severe than hot­water burns. Item 8 An ice cube tray of negligible mass contains 0.280 kg of water at 18.2 ∘ C . How much heat must be removed to cool the water to 0.00 ∘ C and freeze it? Part A Express your answer in joules. ANSWER: Q = 1.15×105 J Correct Part B Express your answer in calories. ANSWER: Q = 27.4 kcal Correct 10/15 8/28/2015 Assignment 1 ­ Chapter 17 Part C Express your answer in Btu. ANSWER: Q = 109 Btu Correct Item 9 The hot glowing surfaces of stars emit energy in the form of electromagnetic radiation. It is a good approximation to assume that the emissivity e is equal to 1 for these surfaces. Part A Find the radius R Rigel of the star Rigel, the bright blue star in the constellation Orion that radiates energy at a rate of 2.7 × 10 31 is spherical. Use σ = 5.67 × 10 −8 W/m 2 ⋅K and has a surface temperature of 11,000 K . Assume that the star W 4 for the Stefan­Boltzmann constant and express your answer numerically in meters to two significant figures. Hint 1. Equation for heat radiation The rate of heat radiation is given by H of the object in kelvins. = AeσT 4 , where A is the surface area of the object, e is the emissivity of the surface, σ is the Stefan­Boltzmann constant, and T is the temperature Hint 2. Surface area of the star Since the star is assumed to be spherical, the surface area is given by A = 4πR 2 , where R is the radius of the star. ANSWER: R Rigel = 5.1×1010 m 11/15 8/28/2015 Assignment 1 ­ Chapter 17 Correct This is over 50 times the size of our own sun and about a third of the orbital radius of the earth around the sun. Rigel is an example of a supergiant star. Part B Find the radius R ProcyonB of the star Procyon B, which radiates energy at a rate of 2.1 × 10 23 W and has a surface temperature of 10,000 K . Assume that the star is spherical. Use σ = 5.67 × 10 −8 W/m 2 ⋅K for the Stefan­Boltzmann constant and express your answer numerically in meters to two significant figures. 4 ANSWER: R ProcyonB = 5.4×106 m Correct This is slightly smaller than the size of the earth and much smaller than (less than 1%) the size of our sun. Procyon B is an example of a white dwarf star. Like many small, dim stars it is visible only through a telescope. Item 10 Two rods, one made of brass and the other made of copper, are joined end to end. The length of the brass section is 0.150 m and the length of the copper section is 0.850 m . Each segment has cross­ sectional area 5.40×10−3 m 2 . The free end of the brass segment is in boiling water and the free end of the copper segment is in an ice and water mixture, in both cases under normal atmospheric pressure. The sides of the rods are insulated so there is no heat loss to the surroundings. Part A What is the temperature of the point where the brass and copper segments are joined? Express your answer to three significant figures and include the appropriate units. ANSWER: T = 61.6 ∘ C 12/15 8/28/2015 Assignment 1 ­ Chapter 17 Correct Part B What mass of ice is melted in 8.80 min by the heat conducted by the composite rod? Express your answer to three significant figures and include the appropriate units. ANSWER: m = 0.238 kg Correct Item 11 A Styrofoam bucket of negligible mass contains 1.75 kg of water and 0.450 kg of ice. More ice, from a refrigerator at −15.0∘ C , is added to the mixture in the bucket, and when thermal equilibrium has been reached, the total mass of ice in the bucket is 0.778 kg. Part A Assuming no heat exchange with the surroundings, what mass of ice was added? Express your answer using three significant figures. ANSWER: m = 0.300 kg Correct Item 12 [ = 2.7 × −5 ∘ ( C −1 [ = 6.8 × −4 ∘ ( C −1 13/15 8/28/2015 Assignment 1 ­ Chapter 17 −1 −1 You are making pesto for your pasta and have a cylindrical measuring cup 11.0 cm high made of ordinary glass [β = 2.7 × 10 −5 (∘ C) ] that is filled with olive oil [β = 6.8 × 10 −4 (∘ C) ] to a height of 1.30 mm below the top of the cup. Initially, the cup and oil are at room temperature (22.0 ∘ C ). You get a phone call and forget about the olive oil, which you inadvertently leave on the hot stove. The cup and oil heat up slowly and have a common temperature. Part A At what temperature will the olive oil start to spill out of the cup? Express your answer using three significant figures. ANSWER: T = 40.3 ∘ C Correct Item 13 You cool a 110.0 g slug of red­hot iron (temperature 745 ∘ C ) by dropping it into an insulated cup of negligible mass containing 75.0 g of water at 20.0 ∘ C . Assume no heat exchange with the surroundings. Part A What is the final temperature of the water? Express your answer using three significant figures. ANSWER: T f inal = 100 ∘ C Correct Part B What is the final mass of the iron and the remaining water? 14/15 8/28/2015 Assignment 1 ­ Chapter 17 Express your answer using three significant figures. ANSWER: m f inal = 181 g Correct Score Summary: Your score on this assignment is 99.0%. You received 98.95 out of a possible total of 100 points. 15/15 ...
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