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Stuck here, help me understand: - Properties of Solids and Liquids - JEE Main-2

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)
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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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