Chapter 16 - Temperature and Heat - Problems and Conceptual Exercises - Page 570: 72

$c_B\lt c_D\lt c_C=c_A$

First of all, we find the specific heat of given materials $c_A=\frac{Q}{(1g)(1K)}=Q J/g/K$ $c_B=\frac{2Q}{(3g)(3K)}=\frac{2}{9}Q J/g/K$ $c_C=\frac{3Q}{(3g)(1K)}=Q J/g/K$ $c_D=\frac{4Q}{(4g)(2K)}=\frac{1}{2}Q J/g/K$ Now we can rank the given materials in increasing specific heat order: $c_B\lt c_D\lt c_C=c_A$