A liquid of mass 250   g is kept warm in a vessel using an electric heater. The liquid is maintained at…

The expression for the rate of heat flow is,

$\frac{\mathrm{d}q}{\mathrm{d}t}=-k\left(T-T_0\right)$

Here $T_0=$Surrounding temperature

From first condition-|

$30=-k\left(57-27\right)$

$\Rightarrow k=-1$

Using second condition-

$\frac{dq}{dt}=-k\left(T-T_0\right)$

$\mathrm{but} \frac{dq}{dt}=ms\frac{dT}{dt}$

$\therefore ms\frac{dT}{dt}=-k \left(T-T_0\right)$

Substitute the given values in the above equation.

$\left(\frac{250}{1000}\right)s\frac{\left(47-46.9\right)}{10}=-\left(-1\right) \left(47-27\right)$

$\Rightarrow s=8000 \mathrm{J} {\mathrm{kg}}^{-1} {\mathrm{K}}^{-1}$