(b) According to Newton’s law of cooling, rate of cooling is given by
$\left( \frac{- dT}{dt} \right) = \frac{eA\sigma}{mc}(T^{4} - T_{0}^{4})$
where c is specific heat of material.
or $\left( \frac{- dT}{dt} \right) \propto \frac{1}{c}$
$ie.,\ $rate of cooling varies inversely as specific heat. From the graph, for A rate of cooling is larger. Therefore, specific heat of A is smaller.