A calorimeter contains 10 g of water at 20⁰C. The temperature falls to 15⁰C in 10 min. When calorimeter contains 20 g of water at 20⁰C, it takes 15 min for the temperature to become 15⁰C. The water equivalent of the calorimeter is

Let water equivalent be x.

$\therefore (10+x)c\frac{\mathrm{\Delta }T}{\mathrm{\Delta }t}=-k\left(\frac{T_1+T_2}{2}-T_0\right)$

where, c is the specific heat of water,

$\mathrm{\Delta }T=\text{ change in temperature }$

and  $\mathrm{\Delta }t=\text{ time at which temperature falls }$

$\therefore (10+x)c\frac{(20-15)}{10}=-k\left(\frac{20+15}{2}-T_0\right)$   …(i)

$\text{ and }(20+x)c\frac{(20-15)}{15}=-k\left(\frac{20+15}{2}-T_0\right)$                …(ii)

From Eqs. (i) and (ii), we get

$\frac{10+x}{10}=\frac{20+x}{15}$

$\Rightarrow 150+15x=200+10x$

$\Rightarrow 5x=50\Rightarrow x=10\mathrm{g}$