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Two identical beakers $A$ and $B$ contain equal volumes of two different liquids at $60\,^oC$ each and left to cool down. Liquid in $A$ has density of $8 \times10^2\, kg / m^3$ and specific heat of $2000\, Jkg^{-1}\,K^{-1}$ while liquid in $B$ has density of $10^3\,kgm^{-3}$ and specific heat of $4000\,JKg^{-1}\,K^{-1}$ . Which of the following best describes their temperature versus time graph schematically? (assume the emissivity of both the beakers to be the same)
Solution
$ – 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)$
$T = {T_0} + \left( {{T_i} – {T_0}} \right){e^{ – kt}}$
$where\,k = \frac{{4e\sigma AT_0^3}}{{ms}}$
$k = \frac{{4e\sigma AT_0^3}}{{\rho vs}}\,\,;\,\,\left| {\frac{{dT}}{{dt}}} \right| \propto k$
$\therefore \left| {\frac{{dT}}{{dt}}} \right| \propto \frac{1}{{\rho s}}$
$\rho A{S_A} = 2000 \times 8 \times {10^2} = 16 \times {10^5}$
${\rho _B}{S_B} = 4000 \times {10^3} = 4 \times {10^6}$
${\rho _A}{S_A} < {\rho _B}{S_B}$
${\left| {\frac{{dT}}{{dt}}} \right|_A} > {\left| {\frac{{dT}}{{dt}}} \right|_B}$
