Two identical beakers A and B contain equal volumes of two different liquids at 60 ° C each and left to cool down. Liquid in A has density of 8 × 10 2 k g m - 3 and specific heat of 2000 J k g - 1 K - 1 while the liquid in B has density 10 3 k g m - 3 and specific heat of 4000 J k g - 1 K - 1 .…

Question

Two identical beakers A and B contain equal volumes of two different liquids at 60 ° C each and left to cool down. Liquid in A has density of 8 × 10 2 k g m - 3 and specific heat of 2000 J k g - 1 K - 1 while the liquid in B has density 10 3 k g m - 3 and specific heat of 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)

Aicharya · Accepted Answer

From Newton's law of cooling,
Rate of cooling:

(- \frac{d T}{d t}) = \frac{4 σ e A T_{0}^{3} (T - T_{0})}{m s}

\propto \frac{1}{m s}

∴ ρ_{A} < ρ_{B}

m_{A} < m_{B}

And

s_{A} < s_{B}

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

At

t = 0, {(- \frac{d T}{d t})}_{A} > {(- \frac{d T}{d t})}_{B}

Initially, the rate of cooling for

A

will be higher than

B

and temperature

(T)

vs time

(t)

graph will be a curve.

Two identical beakers A and B contain equal volumes of two different liquids at 60 ° C each and left to cool…

Solution