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9B-C Physics Assignment #7 Cole [Q17.8] No, you can do work by changing the volume--compressing a gas is one way to change the temperature. Another is to do work against friction--rub your hands together. Likewise, if adding heat goes entirely into doing work, the temperature will not change. [Q17.25] The dark soil will heat very quickly and heat the air layer next to the ground. This hot layer is less dense than the air above it forming an unstable situation. Convection currents will develop in which there will be regions of quickly rising air which will carry the glider higher. [17.5] a) From Eq. (15-1), (9 F/5 C)(40.2 C) + 32 F = 104.4 F, which is cause for worry. b) (9 F/5 C)(12 C) + 32 F = 53.6 F, or 54 F [17.53] The kinetic energy which is converted to heat must raise the temperature to the melting 1 point and then melt the bullet mv 2 = mc + mL 2 v = 2(cT + L) = 2((130 J/kg K)(327.3 C - 25 C) + 24.5 10 3 J/kg) = 357 m/s. [17.70] a) Hc = kc A T (100 C - 65 C) = (385 W/m K)( 4 10 -4 m 2 ) = 5.39 W. x 1m b) If energy is to be conserved, the heat fluxes through both rods must be equal ( energy cannot be 65 C - 0 (100 C - 65 C) created or destroyed any where in the rod. Hc = Hs kc A = ks A 1m L2 k 65 L2 = s = 0.242 m. kc 35 [17.91] a) ds = d d s (1 + s (T - T )) = d d - dos 0.002 in T= + To = + 20 C = 87 C. b) Now both change their length: s dos (1.2 10-5 / C )(2.5 in) ds = d d s (1 + s (T - T )) = d (1 + (T - T )) do s - do -0.002 in T = = -100 C T = T + T = -80 C. = do - do s s (2.0 10 -5 K -1 )(2.502 in )-(1.2 10 -5 K -1 )(2.500 in ) [17.101] a) Any heat taken out of the water will freeze it. The final equilibrium state will be a mixture of ice and water at 0 C. -mwLf + mici (0 - (-10 C)) m c (10 C) 0.075 kg(0.5 cal/gm C )(10 C ) = 4.7 g b) In theory yes, however, a glass mw = i i = Lf 80 cal/g (340 g)(80 cal/g) of water is 12 oz = 340 g which would require mi = = 5.4 kg of ice to (0.5 cal/g C )(10 C) freeze. Obviously, this would not fit in the glass. [17.113] a) Look at where the temperature gradient which drives the conduction. The cold air is just above the top surface of the lake. The sides and bottom are warmer; hence, the heat gets taken out of the layer of water right next to the ice, and it freezes. See diagram. Assigment #7, 11/19/04 Page 1 of 3 UC Davis Air at -10 C x dx WATER at 0 C freezing layer ICE heat flow dm Adx = Adx , dV = density of water and A =surface area of the lake so A dx is the volume of the freezing water dQ dT dt = Hdt = - kA dt . layer. This heat is removed by conduction through the ice: dQ = dt x x t T Equating: - LAdx = - kA dt . Separate and integrate: LA xdx = - kAT dt 0 0 x 1 2 kT LAx 2 = - kATt x = t c) 2 L (920 kg/m 3 )(334 10 3 J/kg) t= (0.25 m )2 = 6.0 10 5 s = 6.9 days o 2(1.6 J/mol K )(10 C) b) To freeze the incremental layer water of requires dQ = - Ldm where dm = d) Using x = 40 m in the above calculation gives t = 1.5 1010 s, about 500 years. 2 [17.116] Note that the 1.5 kW/m is the intensity of the sunlight at the sun 1 earth's orbit where = (av Power)/area. a) Because energy is conserved and cannot be created or destroyed, the power flowing through surface 1 must be the same as the power flowing through surface 2. The earth is r = 1.5 1011m from the sun, so the total power radiated by the sun is Ps = (4r2 ) = (1.5 103W/m2 )(4)(1.5 1011m)2 = 4.2 1026 W!!! At the surface of the sun, Ps 4.2 10 26 W the intensity will be s = = 70 MW/m2 = 4rs 2 4 (6.96 108 m ) H b) Use the Stefan-Boltzmann law: s = = T 4 A 1 2 4 7 10 7 W/m 2 T=4 s = = 5900 K 5.67 10 -8 W/m 2 K 4 Assigment #7, 11/19/04 Page 2 of 3 UC Davis [17.121] a) Assume T2 > T1 , so the flow of heat is radially out. Look at an imaginary spherical surface of radius r where a < r < b. The heat flux or current through the surface is dT H = - kA where A is the area perpendicular to the flow Heat flow dx 2 which is the surface area of the dashed sphere A = 4r By conservation of energy, H at r = a, must be the same as H at a r = r because energy cannot be created or destroyed as it flows r through the shell. H = constant. The intensity of the energy flow must decrease as the heat flows outward because the b surface it is spread over is expanding The temperature gradient is not uniform and must vary with r. dT . Putting it all together, temperature gradient = dr b 1 T1 dT H = - k ( 4r 2 ) Separate variables and integrate: H 2 dr = -4k dT ar T2 dr 4k 4k 4kab H=- (T1 - T2 ) = 1 1 (T2 - T1 ) = (T2 - T1 ) 1 1 b-a - - - b a a b b) To find the temperature of a point inside make the outer limit on the T integral T, and the outer radius on the dr integral r and use the constant nature of H. 4k 4k r - a ab H= (T2 - T ) = 1 1 (T2 - T1 ) T = T2 - (T2 - T1 ) 1 1 ar b - a - - a r a b r-a b T = T2 - (T2 - T1 ) b-a r c) For the cylinder, the heat is flowing through the curved surface of a cylinder so A = 2rL dT . where r is the radius of an imaginary cylindrical surface a < r < b. H = - k (2rL) dr b1 T1 2Lk Integrating and separating, H dr = -2Lk dT H = T -T . b ( 2 1) ar T2 ln a 2Lk 2Lk d) Integrate to T and r instead of T1 and b. H = (T2 - T ) = b (T2 - T1 ) r ln ln a a r ln a T = T2 - (T2 - T1 ) ln b a 4kab 4b 2 e) For a thin spherical shell, let a b and b - a = H = (T2 - T1 ) k (T2 - T1 ). b-a l 2Lk b a+l l l T - T where ln = ln = ln1 + . Because is For a cylindrical shell H = b ( 2 1) a a a a ln a b l l l smal we can expand in a Taylor series, ln = ln1 + ln1 + + Drop the higher order a a a a b l 2Lk 2aL H= T -T k terms, the first term is zero so ln (T2 - T1 ) . b ( 2 1) a a l ln a ( ) ( ) Assigment #7, 11/19/04 Page 3 of 3 UC Davis

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