9/6/2016Chapter 17 Online HW Chapter 17 Online HW Due: 12:45pm on Wednesday, September 7, 2016 To understand how points are awarded, read the Grading Policy for this assignment. Heat versus Temperature The specific heat capacity of aluminum is about twice that of iron. Consider two blocks of equal mass, one made of aluminum and the other one made of iron, initially in thermal equilibrium. Part A Heat is added to each block at the same constant rate until it reaches a temperature of 500 K. Which of the following statements is true? Hint 1. How to approach the problem Heat is added to both blocks at the same constant rate. That is, the same amount of heat is added to each block per unit time. Therefore, the block that reaches the final temperature in the smallest amount of time is the block that requires the smallest amount of heat to undergo the given temperature change. Since both blocks have the same mass and undergo the same temperature change, you can relate the amount of heat absorbed by each block to the block's specific heat capacity. Hint 2. Specific heat capacity Given a sample of mass m of a certain substance, the amount of heat Q needed to change its temperature by an amount ΔT is given by Q = mcΔT , where c is the specific heat capacity characteristic of that substance. It follows that the specific heat capacity c of a sample is the amount of heat required to raise the temperature of one gram of the sample by 1 K. Hint 3. Identify which material requires more heat Consider several onegram samples of different materials. Heat is added to each sample to increase its temperature by 1 K. Which material will absorb the most heat? Hint 1. Definition of specific heat capacity The specific heat capacity c of a sample is the amount of heat required to raise the temperature of one gram of that sample by 1 K. ANSWER: The material with the smallest specific heat capacity will absorb the most heat. The material with the largest specific heat capacity will absorb the most heat. All the materials will absorb the same amount of heat because they all have the same mass. All the materials will absorb the same amount of heat because they all undergo the same change in temperature. ANSWER: The iron takes less time than the aluminum to reach the final temperature. The aluminum takes less time than the iron to reach the final temperature. The two blocks take the same amount of time to reach the final temperature. Correct Part B When the two materials have reached thermal equilibrium, the block of aluminum is cut in half and equal quantities of heat are added to the iron block and to each portion of the aluminum block. Which of the following statements is true? Hint 1. How to approach the problem Since the same quantities of heat are added to samples that have different masses and different specific heat capacities, this may result in different final temperatures for each sample. However, you must keep in mind that each smaller block of aluminum now has half the mass of the https://session.masteringphysics.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=4486905 1/14 but about twice the specific heat capacity. Express your answer in terms of mi . and mi and again to relate ΔTa . Find the temperature change of the smaller aluminum block When an amount of heat Q is absorbed by the block of iron. Hint 2. In this problem the variable is ΔT so write your equations in the form ΔT = ⋯. Write each equation so that all the constants are on one side and the variables are on the other. and m. Find the simplest equation that contains these variables and other known quantities from the problem. Hint 1. The three blocks are no longer in thermal equilibrium; both the aluminum blocks are warmer. Hint 1. Specific heat capacity Given a mass m of a certain substance.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=4486905 2/14 . and Q. and ma . Write this equation twice: once to describe ΔTi . Part A Rank these objects on the basis of their temperatures when removed from the oven. what is its change in temperature ΔTi ? Use mi and ci for the mass of the iron block and the specific heat capacity of iron. Express your answer in terms of mi . ca . ANSWER: ΔTi = Q m i ci Hint 3. compare the two cases presented in the problem. ci .masteringphysics. Finally. the amount of heat Q needed to change its temperature by an amount ΔT is given by Q = mcΔT . and Q. what is its change in temperature ΔTa ? Express the mass of the block in terms of the mass of the iron block mi and the specific heat capacity of aluminum in terms of the specific heat capacity of iron ci . c. To solve this problem use proportional reasoning to find a relation between ΔT . where c is a constant. For this question you should find the ratio ΔTi /ΔTa . respectively.0 kg. ANSWER: ΔTa = Q m i ci ANSWER: The three blocks are no longer in thermal equilibrium; the iron block is warmer. ci .9/6/2016 Chapter 17 Online HW iron block. Find the temperature change of the iron block When an amount of heat Q is absorbed by the block of iron. where the subscript i refers to iron and a to aluminum. Correct Thermal Conductivity Ranking Task Six objects are placed in a 500∘ F (260∘ C) oven and allowed to reach thermal equilibrium. ci . Specific heat capacity of aluminum Recall that the specific heat capacity of aluminum is about twice the specific heat capacity of iron. https://session. The blocks remain in thermal equilibrium. Each object has a mass of 1. The specific heat and thermal conductivity of each substance are denoted by c and k. called specific heat capacity. characteristic of that substance. 9/6/2016 Chapter 17 Online HW Rank from largest to smallest. Rank from largest to smallest. Hint 3. ANSWER: https://session. Equilibrium At equilibrium. if heat energy flows rapidly from your hand when you touch an object. ANSWER: Reset Help largest smallest silver ingot − −−−−−−−− ∘ skillet c = 235iron J/(kg ⋅ C) − − − − − − − − − ∘ ∘ kglass =c 420 = casserole 448 J/(sJ/(kg ⋅ mdish ⋅ ⋅C) C) − −−−−−−−−−−−−−− ∘ ∘ k 837 = 80 J/(s ⋅ ⋅ C) c = J/(kg ⋅ m C) aluminum pot −−−−−−−−− − ∘− ∘ k = 0.2 J/(s ⋅ m ⋅ ∘ C) The correct ranking cannot be determined. Hint 1. Sensing temperature The rate at which heat energy flows into (or out of) your body will determine how hot (or cold) something feels when you touch it. Thermal conductivity The thermal conductivity of a substance is a measure of the rate at which heat energy can flow through a substance. you accidentally touch each one with your hand.8 J/(s ⋅ m ⋅ C) wooden c = 900 cutting J/(kg board ⋅ C) −−−−−−−−−−−−−−−− ∘ − ∘ ⋅ 220 done J/(s ⋅steak m C) ck == 1700 well J/(kg ⋅ C) − −−−−−−−−−−− ∘ ∘ k = 0. an object and its surroundings are at the same temperature.1 J/(sJ/(kg ⋅ m ⋅ ⋅C) c = 3500 C) k = 0. the area A and thickness L of the contacting surface. Correct Part B When removing the objects from the oven. Mathematically. this can be represented as ΔQ Δt =k A L (ΔT ).0 kg of the substance by 1∘ C. A substance with large thermal conductivity allows the rapid flow of heat energy through it. For example. To rank items as equivalent. Rank these objects on the basis of how hot they feel. Hint 1.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=4486905 3/14 . and the thermal conductivity k of the surface. overlap them. between the points. To rank items as equivalent.masteringphysics. overlap them. the object will feel very cold. Specific heat The specific heat of a substance is the amount of heat energy needed to raise the temperature of 1. Hint 2. Hint 2. Rate of heat flow The rate of heat flow (via conduction) ΔQ/Δt from one point to another depends on the temperature difference ΔT . 9/6/2016 Chapter 17 Online HW Reset Help largest smallest silver ingot aluminum pot −−−−−−−−− c = 235 J/(kg ⋅ ∘ −−−−−−−−−−− C) k = 420 J/(s ⋅ m ⋅ ∘ C) c = 900 J/(kg ⋅ ∘ C) k = 220 J/(s ⋅ m ⋅ ∘ C) iron skillet −−−−−−−−− c = 448 J/(kg ⋅ ∘ k = 80 J/(s ⋅ m ⋅ glass casserole dish −−−−−−−−−−−−−−− C) ∘ C) c = 837 J/(kg ⋅ ∘ k = 0.masteringphysics. The tubs are quickly sealed and insulated. Objects that can transfer a large amount of heat energy with a small change in temperature will end up with the highest equilibrium temperatures. Energy transfer and specific heat An object that can transfer a large amount of heat energy with a small change in temperature has a large specific heat. Hint 2. heat energy will flow from the hot objects into the cold water. Conservation of energy Once placed in contact with the cold water. Hint 1. To rank items as equivalent. Rank from largest to smallest.2 J/(s ⋅ m ⋅ C) ∘ C) wooden cutting board −−−−−−−−−−−−−−−−− c = 1700 J/(kg ⋅ ∘ k = 0. overlap them.8 J/(s ⋅ m ⋅ C) ∘ C) well done steak −−−−−−−−−−−− c = 3500 J/(kg ⋅ ∘ k = 0.1 J/(s ⋅ m ⋅ C) ∘ C) The correct ranking cannot be determined.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=4486905 4/14 . lowering the temperature of the objects and raising the temperature of the water. ANSWER: https://session. Correct Part C Each of the objects is immediately dunked in an identical tub of cold water. Rank the objects on the basis of their temperature on reaching equilibrium with the water. 67 × 10 −8 W/m 2 4 ⋅K W and has a surface for the StefanBoltzmann constant and express your answer numerically in meters to two significant figures. Hint 1. where R is the radius of the star. 23 and has a surface temperature of 10. It is a good approximation to assume that the emissivity e is equal to 1 for these surfaces. Assume that the star is spherical.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=4486905 5/14 . which radiates energy at a rate of 2.1 J/(s ⋅ m ⋅ C) ∘ C) aluminum pot glass casserole dish −−−−−−−−−−− c = 900 J/(kg ⋅ ∘ C) k = 220 J/(s ⋅ m ⋅ ∘ C) −−−−−−−−−−−−−−− c = 837 J/(kg ⋅ ∘ k = 0.000 K.2 J/(s ⋅ m ⋅ wooden cutting board −−−−−−−−−−−−−−−−− C) ∘ c = 1700 J/(kg ⋅ C) ∘ k = 0.1×1010 m 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. the bright blue star in the constellation Orion that radiates energy at a rate of 2.000 K. e is the emissivity of the surface. Assume W Use = 5. the surface area is given by A = 4πR 2 . where A is the surface area of the object. and T is the temperature of the object in kelvins. Rigel is an example of a supergiant star.masteringphysics. ANSWER: RRigel = 5. Part B Find the radius RProcyonB of the star Procyon B. Surface area of the star Since the star is assumed to be spherical. Hint 2. https://session. σ is the Stefan Boltzmann constant. Equation for heat radiation The rate of heat radiation is given by H = AeσT 4 .9/6/2016 Chapter 17 Online HW Reset Help largest smallest well done steak −−−−−−−−−−−− c = 3500 J/(kg ⋅ ∘ k = 0. Use σ = 5. Correct The Size of Stars The hot glowing surfaces of stars emit energy in the form of electromagnetic radiation. Part A Find the radius RRigel of the star Rigel.67 × −8 W/ 2 ⋅ 4 for the StefanBoltzmann constant and express your answer numerically in meters to two significant figures.8 J/(s ⋅ m ⋅ ∘ iron skillet −−−−−−−−− c = 448 J/(kg ⋅ C) ∘ k = 80 J/(s ⋅ m ⋅ C) silver ingot −−−−−−−−− C) ∘ C) c = 235 J/(kg ⋅ ∘ C) k = 420 J/(s ⋅ m ⋅ ∘ C) The correct ranking cannot be determined.7 × 10 31 temperature of 11.1 × 10 that the star is spherical. 00 kg of water at 18. After gently stirring for 5.2 Calorimetry. the units of this scale are the same size as those on the Fahrenheit scale (∘ F) rather than the Celsius scale (∘ C).16 K at standard pressure. the Rankine scale is an absolute temperature scale: Absolute zero is zero degrees Rankine (0∘ R).00 minutes.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=4486905 6/14 . what is the specific heat capacity of the metal. Procyon B is an example of a white dwarf star. and vapor) at 273. The specific heat capacity of water is cw = 4190 J/(kg ⋅ K) . liquid. Give your answer to two significant figures. Assuming that the Styrofoam absorbs a negligibly small amount of heat and that no heat was lost to the surroundings. This temperature is known as the triple point.9/6/2016 Chapter 17 Online HW Use σ = 5. ANSWER: RProcyonB = 5.0 ∘ C .0 ∘ C . Rankine Temperature Scale Like the Kelvin scale. Part A Given that water at standard pressure freezes at 0∘ C.00 kg chunk of an unknown metal that has been in boiling water for several minutes is quickly dropped into an insulating Styrofoam beaker containing 1. Triplepoint temperature On the Kelvin temperature scale.8 ∘ F ΔT = 1 K Correct Part B What is the numerical value of the triplepoint temperature Ttriple of water on the Rankine scale? Give your answer to three significant figures.masteringphysics.2 Calorimetry Learning Goal: To practice ProblemSolving Strategy 17. ANSWER: = 1. A 1. The Kelvin temperature scale and the Celsius temperature scale differ only in their zero point. and that it boils at 100∘ C. which corresponds to 212∘ F. calculate the temperature difference ΔT in degrees Fahrenheit that corresponds to a temperature difference of 1 K on the Kelvin scale. Relation of Celsius and Kelvin temperature scales A temperature increase of one kelvin corresponds to a temperature increase of one degree also on the Celsius scale. Like many small.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. cm ? https://session. which corresponds to 32∘ F. ANSWER: Ttriple = 492 ∘ R Correct PSS 17.67 × 10 −8 W/m 2 4 ⋅K for the StefanBoltzmann constant and express your answer numerically in meters to two significant figures. dim stars it is visible only through a telescope. Hint 1. water freezes and coexists in three phases (solid. Hint 1. However. you observe that the water's temperature has reached a constant value of 22. the temperature change for the object is ΔT = Tf − Ti = Tf − 20 ∘ C. a phase change at constant temperature. w m where Qw is the heat gained by the water and Qm is the heat lost by the metal. the amount of heat lost by one body must equal the amount gained by the other body. 4. Qw will be a positive quantity. 3. you may not know in advance whether all the material undergoes a phase change or only part of it. you can think of the metal and the water as two bodies isolated from the surroundings that exchange a certain amount of heat. Heat leaves the water and enters the metal. you will need to find an unknown temperature T . or heat of vaporization. Be certain to identify which quantities are known and which are the unknown target variables. molar heat capacity. Each object will undergo a temperature change with no phase change. SET UP the problem using the following steps Part A In which direction does the heat flow? Hint 1. Solve for the target variables. Often. EXECUTE the solution as follows Part B Assuming that the Styrofoam absorbs a negligibly small amount of heat and that no heat was lost to the surroundings. whereas Qm will be negative. Back up and try again! EVALUATE your answer: A common error is to use the wrong algebraic sign for either a Q or ΔT term. this means that Q +Q = 0. take each quantity of heat added to a body as positive and each quantity leaving a body as negative.9/6/2016 Chapter 17 Online HW ProblemSolving Strategy: Calorimetry problems IDENTIFY the relevant concepts: When heat flow occurs between two bodies that are isolated from their surroundings. 2. 2. the algebraic sum of the quantities of heat transferred to all the bodies must be zero. https://session. cm ? Express your answer in joules per kilogramkelvin. Identify which objects exchange heat. Doublecheck your calculations. You can always assume one or the other. Consult tables for values of the specific heat. or both. ANSWER: Heat enters the metal and enters the water. For example. Use the equation Q = mcΔT to describe temperature changes and equations Q = ±mL f and Q = ±mL v to describe phase changes. what is the specific heat capacity of the metal. To avoid confusion with algebraic signs. heat of fusion. Heat leaves the metal and leaves the water. Heat as energy transfer Heat is the amount of energy that is transferred from one body to another because of a difference in temperature. In problems where a phase change takes place. if an object has an initial temperature of ∘ 20 C and an unknown final temperature Tf . and if the resulting calculation gives an absurd result. In our case. SET UP the problem using the following steps: 1.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=4486905 7/14 .masteringphysics. Note that according to the notation introduced in the strategy. you know the initial assumption was wrong. When two or more bodies interact. EXECUTE the solution as follows: 1. as when ice melts. Correct The principle of conservation of energy for an isolated system requires that the energy that leaves the warmer substance equal the energy that enters the colder substance. IDENTIFY the relevant concepts Since it is assumed that the Styrofoam beaker absorbs a negligibly small amount of heat and that no heat is lost to the surroundings. and make sure that the final results are physically sensible. It always flows spontaneously from an object at higher temperature to one at lower temperature. Calorimetry calculations will allow you to get a quantitative description of the heat exchange between the metal and the water. Heat leaves the metal and enters the water. Express your answer in terms of some or all of the variables mw . respectively.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=4486905 8/14 . cw for the specific heat of water. ANSWER: Q m = m m c m (Tmf − Tmi ) Hint 3. and Twi and Twf for the initial and final temperatures of the water. This suggests that the piece of metal is at the boiling temperature of water when it is dropped into the insulating beaker. This conservation is represented by the equation Qw + Qm = 0.masteringphysics. and Tmi and Tmf for the initial and final temperatures of the metal. Use mm for the mass of the piece of metal. where c is the specific heat capacity of the material. The heat required for a temperature change Recall that the amount of heat Q required to change the temperature of a given material by an amount ΔT is proportional to the mass m of the material and to the temperature change ΔT : Q = mcΔT . Tmi . where c is the specific heat capacity of the material. and Tmf . Hint 1. cm for the specific heat of the metal. Find the heat gained by the water Find a general expression for Qw . Putting it all together Apply conservation of energy: Equate the total heat transferred into and out of the isolated system to zero. The initial and final temperatures of the metal Before being dropped into the Styrofoam beaker. Express your answer in terms of some or all of the variables mm . m w c w ΔTw + m m c m ΔTm = 0 Solve this equation for the unknown cm . The heat required for a temperature change Recall that the amount of heat Q required to change the temperature of a given material by an amount ΔT is proportional to the mass m of the material and to the temperature change ΔT : Q = mcΔT . when the temperature of the water in the insulating beaker has reached a constant value. Substituting the expressions you found for Qw and Qm into this expression yields . Twi . Hint 1. Use mw for the mass of the water. the heat required to change the temperature of the water when it is heated by the metal. the piece of metal has been in boiling water for several minutes. Since no phase changes occurred inside the Styrofoam beaker after the introduction of the piece of metal. and Twf . ANSWER: Q w = m w c w (Twf − Twi ) Hint 2. the metal and the water are in thermal equilibrium. ANSWER: cm = 215 J/(kg ⋅ K) Correct EVALUATE your answer https://session. the heat required to change the temperature of the metal when it is cooled by the water. cm .9/6/2016 Chapter 17 Online HW Hint 1. Hint 4. respectively. cw . Find the heat lost by the metal Find an expression for Qm . 9/6/2016 Chapter 17 Online HW Part C You found that the unknown metal has a specific heat capacity close to. ANSWER: Reset Largest Help Smallest The correct ranking cannot be determined. will experience a greater change in temperature than water when both substances absorb the same amount of heat because it has a lower specific heat capacity. and the temperature change ΔT . Which of the following statements describes this relationship? ANSWER: ΔT is proportional to c. overlap them. the specific heat capacity c of the material. Silver. Correct Your results make sense. for example. PSS 17.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=4486905 9/14 . so it stores more heat per degree of temperature change. To rank items as equivalent. Q = mcΔT . that of silver [c = 234 J/(kg ⋅ K)]. Rank the samples from largest to smallest change in temperature. This is exactly what occurred to the water and the piece of metal placed in the Styrofoam container. Water has one of the highest specific heat capacities of all substances.3: Heat Conduction Learning Goal: https://session. that is. Imagine you have 1 kg of each of the substances listed below. ΔT does not depend upon c. ΔT is inversely proportional to c.masteringphysics. which is a metal. which substance will undergo the largest change in temperature ΔT ? Assume that no substance experiences a phase change. Determine the relationship between the specific heat capacity and the temperature change Recall that the amount of heat Q required to change the temperature of a given material by an amount ΔT is proportional to the mass m of the material. This makes it an ideal substance for hotwater spaceheating systems and other uses that require a minimum drop in temperature for a given amount of heat released into the environment. Hint 1. If the same amount of heat Q is added to each. Let's compare silver to other known materials and see whether your results make sense. you can relate the temperature change to the heat capacity. Since Q and m are the same for each substance in the table. ProblemSolving Strategy 17. then the heat current in both materials must be the same. Often. L EXECUTE the solution as follows: 1. so that some heat flows through each. 4.200 W/(m ⋅ K) ) surrounded by a layer of air trapped inside fur or down. the heat flows through two different materials in succession.50 m in diameter having a layer of fat 4. while the inner surface of the fat layer is at TH = 31. Tangent perpendicular to the radius of the sphere. In this model. In some problems. Assuming the data in the problem are for a steady state.masteringphysics. If heat flows through a single object. it was found that the outer surface layer of the fur is at TC = 2. 3. Direction of heat flow Heat transfer occurs only between regions at different temperatures. If there are two parallel heat flow paths.00×10−2 m thick. but uniform wall thickness. IDENTIFY the relevant concepts In this problem there is heat flow through two different materials in succession. As always. SET UP the problem using the following steps Part A As mentioned in the introduction.0 ∘ C. ask yourself whether the results are physically reasonable. 2. Identify the target variable. EVALUATE your answer: As always. ANSWER: Inward along the radius of the sphere. as shown in the figure. Identify the direction of heat flow in the problem (from hot to cold). https://session.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=4486905 10/14 . SET UP the problem using the following steps 1. so the heat current H must be the same in both materials. in which direction does the heat flow? Hint 1.3 Heat conduction Animals in cold climates often depend on two layers of insulation: a layer of body fat (of thermal conductivity 0. the bear can be modeled as a sphere with different layers. the same heat has to pass through both materials in succession. The temperature T at the interface between the two materials is then intermediate between TH and TC ; In steadystate heat flow. you can approximate it as a flat slab with the same thickness and total wall area.3: Heat conduction IDENTIFY the relevant concepts: The concept of heat conduction comes into play whenever two objects at different temperature are placed in contact. How thick should the air layer (contained within the fur) be so that the bear loses heat at a rate of 50. 2. when a box or other container has an irregular shape.9/6/2016 Chapter 17 Online HW To practice ProblemSolving Strategy 17. use the above equation to solve for the target variable.70 ∘ C during hibernation. and the direction of heat flow is always from higher to lower temperature.40×10−2 W/(m ⋅ K) . then the total H is the sum of the quantities H1 and H2 for the separate paths. We can model a black bear as a sphere 1. In studies of bear hibernation. The temperature at the interface for the fur and fat must be some number between the external and internal temperature of the bear. In the equation H = dQ dt = kA TH −TC L is always measured along that direction and A is always an area perpendicular to that direction. it is essential to use a consistent set of units. Outward along the radius of the sphere.0 W ? Thermal conductivity of air is kair = 2. Hint 1. which is where R is the radius of the sphere.masteringphysics. and an external shell representing the layer of air trapped in the fur.What is the surface area A of the layer of fat? Express your answer in meters squared. the thickness of the layers of insulation is small compared to the radius of the sphere so you can consider the surface area perpendicular to the direction of heat flow as constant. assume that both layers are made of an homogeneous material and that their cross sections perpendicular to the direction of flow are constant. This is reasonable because the thickness of the layers of insulation is small compared to the radius of the sphere. Hint 3. Surface area of a sphere As shown in the figure.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=4486905 11/14 .07 m2 https://session. the thickness and total surface area of the air layer. A = 4πR 2 ANSWER: A = 7. EXECUTE the solution as follows Part B What is the thickness of the air layer (contained within the fur). Equation for heat conduction through the layer of air The heat current H1 through the air layer is H1 = dQ dt = kair A T −TC Lair where kair is the thermal conductivity of the air and Lair and A are. Then you can use these values in the heatcurrent equation to calculate the thickness of the air layer. in the bear model. How to approach the problem Recall that in a steadystate heat flow through two materials in succession. the heat flows along the radius of the sphere. the heat current must be the same in both materials. You must find the temperature at the interface between the fat and the fur. Find the surface area of the fat layer Note that you are modeling the bear as a sphere. respectively. To apply the heat conduction model discussed in the strategy above. Thus. Lair ? Express your answer in meters.9/6/2016 Chapter 17 Online HW Correct By modeling the bear as a sphere. Hint 1. As explained in Part A. You also know the temperature of the outer surface layer of air. Hint 2. the heat current through the fat layer must be the same as the heat current through the layer of air trapped within the fur. the problem translates into the study of heat flow through two spherical shells in succession: an internal shell representing the layer of fat. Your target variable is the thickness of the external layer. The area perpendicular to that direction is the surface area of the sphere. we will assume that also the second layer of insulation has the same surface area A as the layer of fat. kf at . Therefore. is given in W/(m ⋅ K). Make sure to use consistent units for the temperature. Equation for heat conduction through the layer of fat The heat current H2 through the layer of fat is H2 = dQ dt = kf at A TH −T Lf at where kf at is the thermal conductivity of body fat and A and Lf at are. the intermediate temperature at the interface between the two layers doesn't change. you can continue to work using temperatures in degrees Celsius. Hint 1. Note that the heat current is proportional to the temperature gradient (that is. you can equally express the temperature gradient in either ∘ C/m or K/m. is given in W/(m ⋅ K). its numerical value remains unchanged when you express it in ∘ W/(m ⋅ C) because the Celsius and Kelvin scales have the same size unit. respectively.70 ∘ C Correct Even though the thermal conductivity of body fat. TC . Hint 3. the heat currents through the layer of fat and the layer of air trapped in the fur must be equal. Therefore. TH ? Express your answer in degrees Celsius. Find the temperature at the interface between the fat and fur Find the temperature T at the interface between the fat and fur layers. How to approach the problem Because a steady state has been reached. Express your answer in degrees Celsius.masteringphysics. the change in temperature per unit length).6 ∘ C Correct https://session. ANSWER: TH = 31. ANSWER: T = 29. Hint 2. Note that in the steady state. Find the temperature of the outer surface layer of the fur What is the temperature of the outer surface layer of the fur. Express your answers in degrees Celsius. the surface area and thickness of the fat layer. its numerical value remains unchanged when you express it in W/(m ⋅∘ C) because the Celsius and Kelvin scales have the same size unit. Since the Celsius and Kelvin temperature scales have the same size unit. ANSWER: TC = 2. you can continue to work using temperatures in degrees Celsius. set up the heat conduction equation for the fat layer only and solve for the fatfur interface temperature T . kf at . Hint 5.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=4486905 12/14 .0 ∘ C Correct Even though the thermal conductivity of body fat.9/6/2016 Chapter 17 Online HW Correct For the rest of the problem. Thus. Hint 4. Find the temperature at the inner surface of the fat layer What is the temperature at the inner surface of the fat layer. how much work Wf was done on the crate by friction? Use 9. blankets and coats keep us warm in cold weather by trapping the warmer air in regions close to our bodies and hence reducing energy loss by convection and conduction. https://session.4 degrees below the horizontal. Hint 1. 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.9/6/2016 Chapter 17 Online HW ANSWER: L air = 9. ± A Sliding Crate of Fruit A crate of fruit with a mass of 30.12×10−2 m Correct Note that animals' fur serves only as a layer of insulation.8 J Hint 3. it is found that typically in winter black bears have a fur that can grow up to four inches thick or about ten centimeters. Find the difference between initial and final potential energy What is Uf − Ui . To find the work done by friction.40 m/s at the bottom.12 in Correct Your calculations make sense! In fact. Hint 2. Ki Kf = 0. Hint 1. in = 2. apply energy conservation: Ki + Ui + Wnc = Kf + Uf . How to approach the problem If no friction were acting.81 m/s2 for the acceleration due to gravity and express your answer in joules.54 cm ANSWER: L f ur = 9. In a similar way.masteringphysics. EVALUATE your answer Part C How many inches or centimeters thick should the bear's fur layer (Lf ur ) be to hold a layer of air Lair so that adequate insulation is provided? Express your answer using inches (in) or centimeters (cm).80 m down a ramp inclined at an angle of 36. what keeps us warm is not the clothing itself but the air trapped in the clothing. The nonconservative (nc) frictional force is responsible for the difference. The air trapped within the fur has a low thermal conductivity and reduces the transfer of energy away from body due to conduction.87. Part A If the crate was at rest at the top of the incline and has a speed of 2.5 kg and a specific heat capacity of 3550 J/(kg ⋅ K) slides 8. Inches and centimeters Recall that 1 . Find the initial and final kinetic energies What is the kinetic energy of the crate before it starts to slide (Ki ) and after it reaches the bottom of the ramp (Kf )? Express your answer in joules as two terms separated by commas. 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 positions.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=4486905 13/14 . In other words. ANSWER: . In this case Q is a positive quantity since the temperature of the crate is increasing. 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. where m is the object's mass.com/myct/assignmentPrintView?displayMode=studentView&assignmentID=4486905 14/14 .22 m ANSWER: Uf − Ui = 1560 J ANSWER: Wf = 1470 J Correct The frictional force opposes the motion of the crate. c is its specific heat. what is its temperature change ΔT ? Hint 1. ANSWER: h = 5. You received 0 out of a possible total of 0 points. this simplified model provides a useful reminder about the transformation of mechanical energy into thermal energy when nonconservative forces are present.36×10−2 ∘ C Correct Of course. and ΔT is the temperature change (in kelvins) of the object. https://session. the assumptions of "total heat absorption" and "uniform temperature change" are not very realistic; still. What is − Ui = −mgh Express your answer in meters.9/6/2016 Chapter 17 Online HW Hint 1. A helpful formula and a helpful diagram The difference in the potential energy of the crate is given by Uf the value of h? .masteringphysics. Score Summary: Your score on this assignment is 0%. where h is the height shown in the figure below. 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 . ANSWER: ΔT = 1.