question_answer

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 specific heat of\[2000J\,k{{g}^{-1}}{{K}^{-1}}\]while liquid in B has density of \[{{10}^{3}}kg\,{{m}^{-3}}\]and specific heat of \[4000J\,k{{g}^{-1}}{{K}^{-1}}.\] Which of the following best describes their temperature versus time graph schematically? (assume the emissivity of both the beakers to be the same)                                    [JEE Main 8-4-2019 Morning]

A)

B)     

C)

D)     

Correct Answer: D

Solution :

[d] \[-ms\frac{dT}{dt}=e\sigma A\left( {{T}^{4}}-T_{0}^{4} \right)\]

\[-\frac{dT}{dt}=\frac{e\sigma A}{ms}\left( {{T}^{4}}-T_{0}^{4} \right)\]

\[-\frac{dT}{dt}=\frac{4e\sigma AT_{0}^{3}}{ms}\left( \Delta T \right)\]

\[T={{T}_{0}}+({{T}_{i}}-{{T}_{0}}){{e}^{-kt}}\]

where\[k=\frac{4e\sigma AT_{0}^{3}}{ms}\]

\[k=\frac{4e\sigma AT_{0}^{3}}{\rho \text{v}s}\]

\[\left| \frac{dT}{dt} \right|\propto k\]

\[\therefore \]\[\left| \frac{dT}{dt} \right|\propto \frac{1}{\rho s}\]

\[{{\rho }_{A}}{{s}_{A}}=2000\times 8\times {{10}^{2}}=16\times {{10}^{5}}\]

\[{{\rho }_{B}}{{s}_{B}}=4000\times {{10}^{3}}=4\times {{10}^{6}}\]

\[{{\rho }_{A}}{{s}_{A}}<{{\rho }_{B}}{{s}_{B}}\]

\[{{\left| \frac{dT}{dt} \right|}_{A}}>{{\left| \frac{dT}{dt} \right|}_{B}}\]