When an aluminum bar is temporarily connected between a hot reservoir at 790 K a
ID: 2207233 • Letter: W
Question
When an aluminum bar is temporarily connected between a hot reservoir at 790 K and a cold reservoir at 318 K, 2.30 kJ of energy is transferred by heat from the hot reservoir to the cold reservoir. (a) In this irreversible process, calculate the change in entropy of the hot reservoir. (b) In this irreversible process, calculate the change in entropy of the cold reservoir. (c) In this irreversible process, calculate the change in entropy of the Universe, neglecting any change in entropy of the aluminum rod. (d) Mathematically, why did the result for the Universe in part (c) have to be positive?Explanation / Answer
a similar question is solved below, but with different values. hope this helps you. When a system exchanges the heat Q at constant absolute temperature T, its entropy changes by: ?S = Q/T Q (and ?S) is positive, when system absorbs heat, and negative if heat is removed from the system. (a) Heat is absorbed by the cold reservoir. Therefore, ?S_cold = +|Q|/T_cold = 2400J / 298K = 8.055 J/K (b) The same amount of heat is removed from hot reservoir, so ?S_hot = -|Q|/T_hot = -2400J / 830K = -2.892 J/K (c) The change in entropy of the universe equals the the sum of the entropy changes of the reservoirs; ?S_univ = ?S_cold + ?S_hot = 8.055 J/K - 2.892 J/K = 5.163 J/K (d) The reason is that the temperature of the hot reservoir is higher than the temperature. Because the magnitude of heat transferred is the same, the magnitude of the positive change in entropy of the cold reservoir exceeds the negative change in entropy of the hot reservoir. That's why the sum of the entropy changes is positive Formally ?S_univ = ?S_cold + ?S_hot = |Q|/T_cold - |Q|/T_hot = |Q|·(1/T_cold - 1/T_hot) since T_hot > T_cold > 0 (1/T_cold - 1/T_hot) > 0 => ?S_univ > 0
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.