It can be shown that the change in entropy in heating or cooling a sample is given by the relation ΔS = mc ln (T_F/T_i), where m is the mass of the sample, c is the specific heat of the sample, and T_F and T_i are the final and initial temperatures, respectively. What is the change in entropy when 20.0 grams of aluminum with a specific heat of 0.900 J/g K is heated from 10.0°C to 60.0°C? A) +2.93 J/K B) +2.50 J/K C) 0 J/K D) −2.93 J/K

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Quizplus · Accepted Answer

The change in entropy (ΔS) can be calculated using the given formula ΔS = mc ln(T_final/T_initial). Here, m = 20.0 g, c = 0.900 J/g·K, T_initial = 10.0°C = 283.15 K (converting to Kelvin), and T_final = 60.0°C = 333.15 K. Plugging these values into the formula gives ΔS = 20.0 * 0.900 * ln(333.15/283.15) = +2.93 J/K.

It Can Be Shown That the Change in Entropy in Heating