Question

Two identical breakers 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 sp ecific heat o f 2000 J $k{g}^{-1}{K}^{-1}$ while liquid in B has density o f 103 kg m-3 and sp ecific heat o f 4000 J $k{g}^{-1}{K}^{-1}$. 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

Answer: (A)

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