Physics Solution Manual for 1100 and 2101

3 becomes prad e asphere t 4 t04 4 e r 2 t 4

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Unformatted text preview: σ Tbefore A t before Since the same radiant power is emitted before and after the “radiator” is painted, we have Q Q = t after t before or 4 4 eafter σ Tafter A = ebefore σ Tbefore A The terms σ and A can be eliminated algebraically, so this result becomes 4 4 eafter σ Tafter A = ebefore σ Tbefore A or 4 4 eafter Tafter = ebefore Tbefore Remembering that the temperature in the Stefan-Boltzmann law must be expressed in Kelvins, so that Tbefore = 62 °C +273 = 335 K (see Section 12.2), we find that 4 Tafter = 4 ebefore Tbefore eafter or Tafter = 4 ebefore eafter 0.75 (Tbefore ) = 4 0.50 ( 335 K ) = 371 K On the Celsius scale, this temperature is 371 K − 273 = 98 °C . 704 THE TRANSFER OF HEAT 23. REASONING The radiant energy Q absorbed by the person’s head is given by Q = e σ T 4 At (Equation 13.2), where e is the emissivity, σ is the Stefan-Boltzmann constant, T is the Kelvin temperature of the environment surrounding the person (T = 28 °C + 273 = 301 K), A is the area of the head that is absorbing the energy, and t is the time. The radiant energy absorbed per second is Q/t = e σ T 4 A. SOLUTION a. The radiant energy absorbed per second by the person’s head when it is covered with hair (e = 0.85) is ( ) ( ) Q 4 = e σ T 4 A = ( 0.85 ) 5.67 ×10−8 J/ s ⋅ m 2 ⋅ K 4 ( 301 K ) 160 × 10−4 m 2 = 6.3 J/s t b. The radiant energy absorbed per second by a bald person’s head (e = 0.65) is ( ) ( ) Q 4 = e σ T 4 A = ( 0.65 ) 5.67 ×10−8 J/ s ⋅ m 2 ⋅ K 4 ( 301 K ) 160 × 10−4 m 2 = 4.8 J/s t ( ) 24. REASONING The net radiant power of the baking dish is given by Pnet = eσ A T 4 − T04 , (Equation 13.3), where e is the emissivity of the dish, σ is the Stefan-Boltzmann constant, A is the total surface area of the dish, T is the Kelvin temperature of the dish, and T0 is the temperature of the kitchen. As the dish cools down, both its net radiant power Pnet and temperature T decrease, but its emissivity e and surface area A remain constant. We will therefore solve Equation 13.3 for the product of these constant quantities and the StefanBoltzmann constant: Pnet eσ A = (1) 4 T − T04 ( ) Because the left side of Equation (1) is constant, the ratio on the right hand side cannot change as the temperature T decreases. We will use this fact to determine the radiant power Pnet,1 of the baking dish when its temperature is 175 °C. SOLUTION From Equation (1), we have that Pnet,1 = Pnet,2 (T14 − T04 ) (T24 − T04 ) or T 4 − T 4 Pnet,1 = Pnet,2 14 04 T2 − T0 (2) In Equation (2), Pnet,2 = 12.0 W is the net radiant power when the temperature is 35 °C. We first convert temperatures to the Kelvin scale by adding 273 to each temperature given in degrees Celsius (see Equation 12.1). The initial temperature of the baking dish is Chapter 13 Problems 705 T1 = 175 °C + 273 = 448 K, its final temperature is T2 = 35 °C + 273 = 308 K, and the room temperature is T0 = 22 °C + 273 = 295 K. Equation (2), then, gives the net radiant power when the dish is first brought out of the oven: ( 448 K )4 − ( 295 K )4 = 275 W Pnet,1 = (12.0 W ) ( 308 K )4 − ( 295 K )4 25. REASONING According to the discussion in Section 13.3, the net power Pnet radiated by ( ) the person is Pnet = eσ A T 4 − T04 , where e is the emissivity, σ is the Stefan-Boltzmann constant, A is the surface area, and T and T0 are the temperatures of the person and the environment, respectively. Since power is the change in energy per unit time (see Equation 6.10b), the time t required for the person to emit the energy Q contained in the dessert is t = Q/Pnet. SOLUTION The time required to emit the energy from the dessert is t= Q Q = Pnet eσ A T 4 − T 4 0 ( ) 4186 J The energy is Q = ( 260 Calories ) , and the Kelvin temperatures are 1 Calorie T = 36 °C + 273 = 309 K and T0 = 21 °C + 273 = 294 K. The time is t= ( 260 Calories ) ( 0.75 ) 5.67 ×10−8 4186 J 1 Calorie = 1.2 ×104 s 4 4 2 4 2 J/ s ⋅ m ⋅ K 1.3 m ( 309 K ) − ( 294 K ) ( )( ) 26. REASONING The power radiated by an object is given by Q / t = eσ T 4 A (Equation 13.2), where e is the emissivity of the object, σ is the Stefan-Boltzmann constant, T is the temperature (in kelvins) of the object, and A is its surface area. The power that the object absorbs from the room is given by Q/t = eσT0 A. Except for the temperature T0 of the room, this expression has the same form as that for the power radiated by the object. Note especially that the area A is the surface area of the object, not the room. Review Example 8 in the text to understand this important point. 4 706 THE TRANSFER OF HEAT SOLUTION The object emits three times as much power as it absorbs from the room, so it follows that (Q/t)emitted = 3(Q/t)absorbed. Using the Stefan-Boltzmann law for each of the powers, we find eσ T 43 3e 4 A 1 24 = 1 σ T0 3 4A 4 24 Power emitted Power absorbed Solving for the temperature T of the object gives T = 4 3T0 = 4 3 ( 293 K ) = 386 K 27. SSM REASONING AND SOLUTION The net power generated by the stove is given by Equation 13.3, Pnet = e...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.

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