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Two identical beakers A and B contain equal volumes of two different liquids at $60^{o}C$ each and left to cool down. Liquid in A has density of $8\times 10^{2}kg/m^{3}$ and specific heat of $2000\; J\; kg^{-1}K^{-1}$ while liquid in B has density of $10^{3}\; kg\: m^{-3}$ and specific heat of $4000\; J\; kg^{-1}K^{-1}$ which of the following best describes their temperature versus time graph schematically ?(assume the emissivity of both the beakers to be the same)
Option 1)
Option 2)
Option 3)
Option 4)

Answers (1)

best_answer

From given

$\rho _{A}< \rho _{B},$

$\Rightarrow m _{A}<m _{B},$

and $\Rightarrow s _{A}<s _{B},$

By newtons law of cooling

$\frac{-dT}{dt}=\frac{4\: \sigma\: e\: A\: T{_{0}}^{3}(T-T_{o})}{ms}$

$\Rightarrow \frac{-dT}{dt}\alpha \frac{1}{ms}$

at $t=0$

$-(\frac{dT}{dt})_{A}\alpha \frac{1}{m_{A}s_{A}}$

$-(\frac{dT}{dt})_{B}\; \alpha \frac{1}{m_{B}s_{B}}$

and we know

$m_{A}s_{A}<m_{B}s_{B}$

So slope of T v/s t curve for A is more than B.

Option 1)

Option 2)

Option 3)
Option 4)

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