Question

Two identical beakers $A$ and $B$ contain equal volumes of two different liquids at ${60}^{\circ }C$ each and left to cool down. Liquid in $A$ has density of $8\times {10}^{2}kg/m^3$ and specific heat of $2000Jk{g}^{-1}{K}^{-1}$ while liquid in $B$ has density of ${10}^{3}kg{m}^{-3}$ and specific heat of $4000Jk{g}^{-1}{K}^{-1}$. Which of the following best describes their temperature versus time graph schematically? (assume the emissivity of both the beakers to be the same?

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

$-ms\frac{dT}{dt}=e\sigma A(T^4-T_0^4)$
$-\frac{dT}{dt}=\frac{e\sigma A}{ms}(T^4-T_0^4)$
$-\frac{dT}{dt}=\frac{4e\sigma A{T}_0^3}{ms}\left(△T\right)$
$T=T_0+(T_i-T_0)e^{-kt}$
where $k=\frac{4e\sigma A{T}_0^3}{ms}$
$k=\frac{4e\sigma A{T}_0^3}{\rho vs}$
$\left|\frac{dT}{dt}\right|\propto k$
$\therefore \left|\frac{dT}{dt}\right|\propto \frac{1}{\rho s}$
${\rho }_A{S}_A=2000\times 8\times {10}^2=16\times {10}^5$
${\rho }_B{S}_B=4000\times {10}^3=4\times {10}^6$
${\rho }_A{s}_A<{\rho }_B{s}_B$
${\left|\frac{dT}{dt}\right|}_A>{\left|\frac{dT}{dt}\right|}_B$.