A calorimeter with negligible heat capacity holds 100 cc of water...

According to Newton's law of cooling, the rate of heat loss is proportional to the temperature difference between the liquid and the surroundings.

The rate of heat loss is given by $\dfrac{dQ}{dt} = ms \dfrac{dT}{dt}$.

Since the temperature range and surrounding conditions are identical for both liquids, the average rate of heat loss is the same.

Therefore, the time taken to cool is proportional to the total heat capacity of the liquid.

$t \propto ms$

Since mass $m = V \rho$, we have $t \propto V \rho s$.

Given that the volumes are equal, the ratio of the times taken is:

$\dfrac{t_2}{t_1} = \dfrac{\rho_2 s_2}{\rho_1 s_1}$

Substituting the given values:

$\dfrac{t_2}{5} = \dfrac{800 \times 2100}{1000 \times 4200}$

$\dfrac{t_2}{5} = \dfrac{8}{10} \times \dfrac{1}{2} = 0.4$

$t_2 = 5 \times 0.4 = 2 \text{ min}$