Acopper block of 500 g with a temperature of 293 K is brought intothermal contac

Question

Acopper block of 500 g with a temperature of 293 K is brought intothermal contact with a resistive heater of 1000 andnegligible mass. For 15 s, a current of 1.0 A is passed through theheater. (a) What is the final temperature of the copper block? (b)By how much does the entropy of the copper block change? Assume cp= 24.4 J mol-1 K-1 (constant). The block is then placed into athermally insulated container and 134 g of water with a temperatureof 25°C is poured over it. (c) What is the final temperature ofthe block and the water? (d) What is the total increase in entropyduring the thermal equilibration?

Explanation / Answer

a)   Power = i^2*R = (1A)2 *(1000 ohm) = 1000W Energy = Power * time = 1000 W* 15 s = 15000 Joules delU = Q + W = Q b/c no work delU = Q = mass*C*T T = Energy/ [mass*C] = [15000 J] / [ 500 g * 1 mol / 63.54g * 24.4 J/mol/K] = 78.1229508 K Tfinal = Tinit + T = 293 K + 78.12 K = 371.122951 = 371K b) S = mass*integral [ Cv*dT/T] = Cv*ln(Tfinal/ Tinitial) = (500g * 1 mol/63.54 g * 24.4 J/mol/K)*ln(371 K/ 293K) S = 45.3188438 J/K S = 45.3 J/K c) Qreleased by copper = Q absorbed by water (500 g * 1mol/63.54 g* 24.4 J/mol/K) * (371 K - Tfinal) = 134 g*(4.18 J/g/C) *( Tfinal - 298K) 192.005036 *(371 - Tfinal) = 560.12 *(Tfinal - 298K) Tfinal = (192*371 + 560*298)/(560 + 192) Tfinal = 316.638298 K ~ 317 K d) S = masscopper*C*ln(Tf/Ti) + mass water*C*ln(Tf/Ti) S = (500 g * (1mol/63.54 g* 24.4 J/mol/K))* ln(317 K / 371K)+ 134 g *(4.18 J/g/K)*ln(317 K / 298 K) S == 4.41761029 J/K = 4.42 J/K

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