Solution
According to Newton’s Law of Cooling, the rate of cooling is given by:
$$ -\frac{dQ}{dt} = K(T – T_0) $$
Writing this in terms of temperature change $dT/dt$:
$$ (ms)_{total} \frac{dT}{dt} = -K(T – T_0) \implies dt = \frac{(ms)_{total}}{K(T-T_0)} dT $$
The time $t$ taken to cool between the same temperature limits is proportional to the total heat capacity $(ms)_{total}$ of the system (assuming the geometry and surface area $K$ remain constant).
Case 1 (Water only):
Let mass of water be $m$ and specific heat be $s_w$.
$t \propto m s_w$
Case 2 (Water + Ball):
Mass of ball is equal to mass of water ($m$). Specific heat of ball is $s_b$.
$t’ \propto (m s_w + m s_b)$
Given $t’ = k \times t$.
Taking the ratio:
$$ \frac{t’}{t} = \frac{m s_w + m s_b}{m s_w} $$
$$ k = \frac{s_w + s_b}{s_w} = 1 + \frac{s_b}{s_w} $$
$$ k – 1 = \frac{s_b}{s_w} \implies s_b = (k-1)s_w $$
Correct Option: (c) $ s_b = (k-1)s_w $