536_PartUniversity Physics Solution

536_PartUniversity Physics Solution - 17-22 Chapter 17...

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17-22 Chapter 17 17.110. IDENTIFY: Apply Eq.(17.23). SET UP: Let 1 1 HR T A Δ= be the temperature difference across the wood and let 2 2 HR T A be the temperature difference across the insulation. The temperature difference across the combination is 12 . TTT Δ+ Δ The effective thermal resistance R of the combination is defined by . HR T A EXECUTE: Δ gives (), HH R RR AA += and . R =+ EVALUATE: A good insulator has a large value of R . R for the combination is larger than the R for any one of the layers. 17.111. IDENTIFY and SET UP: Use H written in terms of the thermal resistance R : / , H ATR = Δ where / R Lk = and RRR =++ (additive). EXECUTE: single pane s glass film , RR R where 2 film 0.15 m K / W R =⋅ is the combined thermal resistance of the air films on the room and outdoor surfaces of the window. 32 glass / (4.2 10 m)/(0.80 W/m K) 0.00525 m K / W RL k ==× = Thus 22 2 s 0.00525 m K / W .15 m K / W 0.1553 m K / W. R + = double pane dg l a s s a i r f i l m 2, R R + where air R is the thermal resistance of the air space between the panes. air / (7.0 10 m)/(0.024 W/m K) 0.2917 m K / W k = Thus 2 2 d 2(0.00525 m K / W) 0.2917 m K / W 0.15 m K / W 0.4522 m K / W R + + = ss dd s d d s /, s o / / H ATR H ATR H H R R = (since A and T Δ are same for both) sd / (0.4522 m K / W)/(0.1553 m K / W) 2.9 = EVALUATE: The heat loss is about a factor of 3 less for the double-pane window. The increase in R for a double- pane is due mostly to the thermal resistance of the air space between the panes. 17.112. IDENTIFY: kA T H L Δ = to each rod. Conservation of energy requires that the heat current through the copper equals the sum of the heat currents through the brass and the steel. SET UP: Denote the quantities for copper, brass and steel by 1, 2 and 3, respectively, and denote the temperature at the junction by 0 . T EXECUTE: (a) 123 . H Using Eq.(17.21) and dividing by the common area gives, () 3 00 0 3 100 C . kk k TT T LL L °− = + Solving for 0 T gives ( ) 11 0 2 2 33 100 C . kL T = ° ++ Substitution of numerical values gives 0 78.4 C. T (b) Using kA H T L for each rod, with 3 21.6 C , 78.4 C T °Δ= ° gives 12.8 W, 9.50 W == and 3 3.30 W. H = EVALUATE: In part (b), 1 H is seen to be the sum of 23 and . H H 17.113. (a) EXECUTE: Heat must be conducted from the water to cool it to 0&C and to cause the phase transition. The entire volume of water is not at the phase transition temperature, just the upper surface that is in contact with the ice sheet. (b) IDENTIFY: The heat that must leave the water in order for it to freeze must be conducted through the layer of ice that has already been formed. SET UP: Consider a section of ice that has area A. At time t let the thickness be h. Consider a short time interval t to . td t + Let the thickness that freezes in this time be dh. The mass of the section that freezes in the time interval dt is . dm dV A dh ρ The heat that must be conducted away from this mass of water to freeze it is ff . dQ dmL AL dh /( / ) , H dQ dt kA T h Δ so the heat dQ conducted in time dt throughout the thickness h that is already there is HC . dQ kA dt h ⎛⎞ = ⎜⎟ ⎝⎠ Solve for dh in terms of dt and integrate to get an expression relating h and t .
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Temperature and Heat 17-23 EXECUTE: Equate these expressions for dQ.
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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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536_PartUniversity Physics Solution - 17-22 Chapter 17...

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