Solution
Both liquids cool over the same temperature range ($50^\circ\text{C}$ to $45^\circ\text{C}$) in identical calorimeters. The time taken ($t$) is proportional to the total thermal capacity of the system.
$$ t \propto (m_{liquid}s_{liquid} + m_{cal}s_{cal}) $$
Data:
$t_w = 60 \text{ s}$ (Water)
$t_l = 40 \text{ s}$ (Liquid)
Masses are equal: $m_w = m_l = m_{cal} = m$
$s_w = 4.20 \text{ J/g}^\circ\text{C}$
$s_{cal} = 0.36 \text{ J/g}^\circ\text{C}$
Let $s_l$ be the specific heat of the unknown liquid.
Ratio Equation:
$$ \frac{t_w}{t_l} = \frac{m(s_w) + m(s_{cal})}{m(s_l) + m(s_{cal})} $$
$$ \frac{60}{40} = \frac{4.20 + 0.36}{s_l + 0.36} $$
$$ 1.5 = \frac{4.56}{s_l + 0.36} $$
$$ s_l + 0.36 = \frac{4.56}{1.5} = 3.04 $$
$$ s_l = 3.04 – 0.36 = 2.68 \text{ J/g}^\circ\text{C} $$
Correct Option: (b) 2.68