It can be shown that the change in entropy in heating or cooling a sample is given by the relation $\Delta$S = 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 grams of aluminum with a specific heat of 0.900 J/gm K is cooled from 60.0$^{\circ}$C to 10.0$^{\circ}$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

Given: m = 20 g c = 0.900 J/gm K T_F = 10.0 °C = 283.15 K T_i = 60.0 °C = 333.15 K Using the formula: ΔS = mc ln (T_F/T_i) ΔS = (20 g) (0.900 J/gm K) ln (283.15 K/333.15 K) ΔS = -2.93 J/K The negative sign indicates that the process is not spontaneous, and the entropy of the system decreases. Hence the correct option is D.

It Can Be Shown That the Change in Entropy in Heating