Question

5) A 3.50 kg piece of lead and a 1.5 kg piece of aluminum are placed in contact. They are able to exchange energy as heat, and they can only exchange it with each other. The initial temperature of the lead is 48°C, and the initial temperature of the aluminum is 35°C. a) If it is possible, construct a single, closed Energy-Interaction diagram for the process of lead and aluminum coming to thermal equilibrium. Write an algebraic expression of energy conservation in terms of the AEs of the two energies, and then substitute in the appropriate expressions for the AEs. Finally, substitute in all known numerical values. Do you have enough information to find the equilibrium temperature? b) At an intermediate point in the entire process described above the lead has reached a temperature of 43°C. Follow the instructions for part (a) for the interval that ends when the lead is at a temperature of 43°C. Determine the unknown(s) in your equation.

Answer 1

. m. lead m Specific heat 2) 3.5 kg Al = 1.5kg lead ce 0.16 kJ/kg of Cai =) 0.9 kilky Tai = 35°c Te=48°C lead head energy = heat energy lost by lead AE, -Alie gained by Al so l AE (1) C, AT so me ATA - m Al Al CAT Al (48-74) Te =) 305 x 0.16 x 18 x (T1-35)= 1.5 toqx78x ») 0.4148*(T7 -35) = 48-74 0-4148.17 - 14:518 48-TH 1-4148 = 62.518 TA .62.518 1-4148 Final temperatre Tf 3 44.188 °C

» At an intermediate state 43°C 4892 Tylum = 35 °c Tleada Tload i I Aluna ? energy balace again so apply again at this instant ДЕ AE Loss of bead gain by Al 3.5xo 0.16 xhé! (481-43) = 1.5*0.9xJ8 xl Tour -35 ТА , 35+2.07 2 Tala 37-07 C or e