541_PartUniversity Physics Solution

# 541_PartUniversity Physics Solution - Temperature and Heat...

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Temperature and Heat 17-27 EVALUATE: A plot of temperature as a function of both position and time for 0 50 s t is shown in Figure 17.124c. Figure 17.124

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17-28 Chapter 17 17.125. IDENTIFY: Apply the concept of thermal expansion. In part (b) the object can be treated as a simple pendulum. SET UP: For steel 51 1.2 10 (C ) . α −− ° 1 yr 86,400 s. = EXECUTE: (a) In hot weather, the moment of inertia I and the length d in Eq.(13.39) will both increase by the same factor, and so the period will be longer and the clock will run slow (lose time). Similarly, the clock will run fast (gain time) in cold weather. (b) 4 0 (1.2 10 (C ) )(10.0 C ) 1.2 10 . L T L Δ =Δ= × ° °= × (c) See Problem 13.98. To avoid possible confusion, denote the pendulum period by . τ For this problem, 5 1 6.0 10 2 L L Δ Δ == × so in one day the clock will gain 5 (86,400 s)(6.0 10 ) 5.2 s. ×= (d) 1 2 . T Δ 1.0 s 86,400 s Δ = gives 1 2[(1.2 10 (C ) )(86,400)] 1.9 C . T Δ ° = ° T must be controlled to within 1.9 C °. EVALUATE: In part (d) the answer does not depend on the period of the pendulum. It depends only on the fractional change in the period. 17.126. IDENTIFY: The rate at which heat is absorbed at the blackened end is the heat current in the rod, 44 S2 2 1 () ( ) kA Ae T T T T L σ −= where 12 20.00 K and TT = is the temperature of the blackened end of the rod. SET UP: Since the end is blackened, 1. e = s 500.0 K. T = EXECUTE: If the equation were to be solved exactly for 2 , T the equation would be a quartic, very likely not worth the trouble. Following the hint, approximate 2 T on the left side of the above expression as T 1 to obtain S 24 1 23 21 1 1 s 1 1 ( ) (6.79 10 K )( ) 0.424 K 20.42 K. L TT T k = + + × + = EVALUATE: This approximation for 2 T is indeed only slightly than 1 , T and is a good estimate of the temperature. Using this for 2 T in the original expression to find a better value of T Δ gives the same T Δ to eight figures, and further iterations are not worthwhile. 17.127. IDENTIFY: The rate in (iv) is given by Eq.(17.26), with 309 K T = and s 320 K. T = The heat absorbed in the evaporation of water is . Qm L = SET UP: , mV ρ = so . m V = EXECUTE: (a) The rates are: (i) 280 W, (ii) 22 (54 J/h C m )(1.5 m )(11 C )/(3600 s/h) 0.248 W, ⋅° ° = (iii) 3 (1400 W /m) ( 1 .5 m) 2 .10 10 W , (iv) 82 4 2 4 4 (5.67 10 W/m K )(1.5 m )((320 K) (309 K) ) 116 W. ×⋅ = The total is 2.50 kW, with the largest portion due to radiation from the sun. (b) 3 63 36 v 2.50 10 W 1.03 10 m /s. (1000 kg m )(2.42 10 J kg K) P L × × This is equal to 3.72 L/h. = (c) Redoing the above calculations with 0 e = and the decreased area gives a power of 945 W and a corresponding evaporation rate of 1.4 L/h. Wearing reflective clothing helps a good deal. Large areas of loose weave clothing also facilitate evaporation.
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## This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.

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541_PartUniversity Physics Solution - Temperature and Heat...

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