Question

Chapter 20, Problem 007 A 22.4 g block of copper whose temperature is 369 K is placed in an insulating box with a 80.89 block of lead whose temperature is 242 K. (a) What is the equilibrium temperature of the two-block system? (b) What is the change in the internal energy of the two-block system between the initial state and the equilibrium state? (c) What is the change in the entropy of the two-block system? The heat capacities of copper and lead are 386 W/kg-K and 128 1/kg.K, respectively. (a) Number Units (b) Number Units (c) Number Units

Answer 1

a Mass of copper block m, = 22-4 g -3 I g = 10 3 Kg mi = 22.4 x 10 Kg = -3 m = 80.8 g - 2 80.8 X10 Kg - 3 - Initial temperature of copper block T, = 369k Mass of lead block = Initial temperature of lead block T2 =242K Let equilibrium temperature of of two block system T (9) Now, to reach equilibrium heat lost lost by by higher temperature body is equal heat gained by lower temperature body. Here copper block has higher temperature than lead block Now, Heat lost by copper = Q = mici CT, -T) Where C, is heat capacity of copper heat capacity of copper = 386 J /kg-k Now, Heat gained by lead block = m2 C₂ CT-T2) / to © OG where Cz 2 heat capacity of lead hare Now, at equilibrium , we = 128 J/kg-K a (0 (0 L 3 =) - m2 C2 mic, (Ti-T) (T-T2) - 22.4 x 10 Kg X 386 J 1 Kg -K (T, - 1) = 80.8810 kg x 12.8 Jl Kg kg x 128 J/kg-K (T-T2) -3 -3 5 => 22.4 x386 (T; - T) = 80.8 X 128 (T-T2) 8646.4 (Ti-T) = 10342.4 (T-T2) 10342.4 Ti-T (T-T2) 8646:ų =)

2 T, - T = 1. 196151 (T =) Ti-T.= 1.19 6151 T Tz) 1.196151 Tz - -) T + 1.19 6151 Tz 1.19 615 IT +T = T +1.196151 Tz 2.19615 IT T + 1.19 6151 Tz T 2.19615 l -) (369 k) + 1.796151 ( 242 K) - T = 2.196151 369 K + 289.468542k 2.196151 658.468542K =) = T = 2.) 9 6 15 , tuo =) T = 2 99.828 4 4 K ñ 300K block system Hence, equilibrium temperature of il 300k b) Take heat gain as positive and heat lost negative Then Net heat change of system is da = On-QL =) do QG-au (using ) as da = 0 Now, box is insulated volume change of system is 30 / also zero . l-e dv=o = Work done Pressure x change in volume dw Pressure xdv dw= Pressure Xo dw so Now, to first law of thermodynamics we have 4 du = du td w where : du is change in internal energy of the system according

3 From 9 We hare dU = d Q - dw du = du so change in internal of tuo block system between initial and equilibrium states is zero(o). so change energy a (C) Let temperature of substance of mass m and heat capacity C is change by small temperature d T. Then change in heat of substance do=mc d T Let any instant temperature is T, then change in entropy is da mcdT 16 ds ds ㅜ T S at any = ت change entropy of Sl Tf = T Let temperature changes from Ti to Tf then in is giren by (Integrating ) Sf Tf mcdT sas das mcl dt T si si Ti where s and St entropies at temperatures Ti and Tf respectively. Iff mc [ent], Si T are Tf [s] Ti = - ک - مک = MC لا en Tf -In Ti =) AS mc en en ( m) [: en a • enb en l (9)] 기

า So, by by using Change in enteroby block is ASC of copper Т. en T mici Change in enteropy of lead block is AS, = m2 Cz en (7) 9 is AS = So, total change in entropy of tuo block system ASC TAS, m, C, en (5) + m2 (2 en ( 17 / 2 $=(22.4 x 1 2.48103 kg x 386 J Kyll [80.8 xloº kg X 128 J Kg?k x en 299.82844K AS x en 1299.828 44K "K) 36 g K -3 + 8X1 242 k -) =) As = [8.6464 JK'x en (0.8125412)) + [10.34240 JK” xen (1.23896)] =) as= [8.6464 J K [8.6464 J K ' X (-0.2075862)] +[10034240 J K X (0.2142723)] 1.794g J K + 2.2161 JK 0.4212 J. K 0:421 JK) So, change of two block 0.421 J Ik As - =) Δς in entropy system is