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 cooled from 60.0°C to 10.0°C? A) +2.93 J/K B) +2.50 J/K C) 0.00 J/K D) −2.93 J/K

Question

Quizplus · Accepted Answer

First, we need to convert the mass of the sample to kilograms: m = 20.0 g = 0.0200 kg Then we can use the given formula to calculate the change in entropy: ΔS = mc ln (T_F/T_i) ΔS = (0.0200 kg)(0.900 J/g K) ln (333 K/283 K) ΔS = -2.93 J/K Note that the result is negative because the sample is being cooled (entropy is decreasing). Therefore, the correct answer is D.

It Can Be Shown That the Change in Entropy in Heating