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Unformatted text preview: CHAPTER 13 THE TRANSFER OF HEAT ANSWERS TO FOCUS ON CONCEPTS QUESTIONS 1. (d) The heat conducted during a time t through a bar is given by ( 29 , k A T t Q L ∆ = where k is the thermal conductivity, and A and L are the crosssectional area and length of the bar. 2. (b) This arrangement conducts more heat for two reasons. First, the temperature difference ∆ T between the ends of each bar is greater in A than in B. Second, the crosssectional area available for heat conduction is twice as large in A as in B. A greater crosssectional area means more heat is conducted, everything else remaining the same. 3. (e) This arrangement conducts more heat for two reasons. First, the crosssectional area A available for heat flow is twice as large in B than in A. Twice the crosssectional area means twice the heat that is conducted. Second, the length of the bars in B is onehalf the combined length in A, which also means that twice the heat is conducted in B as in A. Thus, the heat conducted in B is 2 × 2 = 4 times greater than that in A. 4. (e) The heat conducted through a material is given by ( 29 k A T t Q L ∆ = . The heat Q , crosssectional area A , thickness L , and time t are the same for the three materials. Thus, the product k ∆ T must also be the same for each. Since the temperature difference ∆ T across material 3 is less than that across 2, k 3 must be greater than k 2 . Likewise, since the temperature difference across material 2 is less than that across 1, k 2 must be greater than k 1 . 5. k 2 = 170 ( 29 J/ s m C × × ° 6. (b) The heat conducted through a material is given by ( 29 k A T t Q L ∆ = (Equation 13.1). The heat Q , crosssectional area A , length L , and time t are the same for the two smaller bars. Thus, the product k ∆ T must also be the same for each. Since the thermal conductivity k 1 is greater than k 2 , the temperature difference ∆ T across the left bar is smaller than that across the right bar. Thus, the temperature where the two bars are joined together (400 ° C  ∆ T ) is greater than 300 ° C. 7. (b) The eagle is being lifted upward by rising warm air. Convection is the method of heat transfer that utilizes the bulk movement of a fluid, such as air. 175 THE TRANSFER OF HEAT 8. (c) The radiant energy emitted per second is given by 4 / Q t e T A σ = (Equation 13.2). Note that it depends on the product of T 4 and the surface area A of the cube. The product T 4 A is equal to 4 2 1944 T L , 4 2 1536 T L , and 4 2 864 T L for B, A, and C, respectively. 9. Energy emitted per second = 128 J/s 10. (a) The radiant energy emitted per second is given by 4 / Q t e T A σ = (Equation 13.2). The energy emitted per second depends on the emissivity e of the surface. Since a black surface has a greater emissivity than a silver surface, the blackpainted object emits energy at a greater rate and, therefore, cools down faster....
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This note was uploaded on 03/30/2012 for the course PHYSICS 201 taught by Professor Rollino during the Fall '11 term at Rutgers.
 Fall '11
 rollino
 Physics, Heat

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