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 20.0 grams of aluminum with a specific heat of 0.9 J/gm K is heated from 10.0$^{\circ}$C to 60.0$^{\circ}$C? A) +2.93 J/K B) +2.50 J/K C) 0 J/K D) -2.93 J/K

Question

Quizplus · Accepted Answer

The change in entropy 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.Substituting the given values, we get:$\Delta$S = (20.0 g)(0.9 J/g K) ln (60.0^$\circ$C/10.0^$\circ$C) = +2.93 J/KTherefore, the change in entropy is +2.93 J/K.

It Can Be Shown That the Change in Entropy in Heating