JEE PYQ: Thermodynamics Question 31
Question 31 - 2019 (08 Apr Shift 1)
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 $8 \times 10^2$ 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)
(1) A cools faster (steeper curve for A)
(2) B cools faster (steeper curve for B)
(3) Both cool at same rate (same curve)
(4) A and B cool together (single curve labeled “A and B”)
Show Answer
Answer: (2)
Solution
Rate of Heat loss $= ms\left(\frac{dT}{dt}\right) = e\sigma AT^4$
$\frac{dT}{dt} = \frac{e\sigma \times A \times T^4}{\rho \times Vol \times S} \Rightarrow \frac{dT}{dt} \propto \frac{1}{\rho S}$
$\frac{\left(\frac{dT}{dt}\right)_A}{\left(\frac{dT}{dt}\right)_B} = \frac{\rho_B \times S_B}{\rho_A \times S_A} = \frac{10^3}{8 \times 10^2} \times \frac{4000}{2000} = 2.5 > 1$
$\Rightarrow \left(\frac{dT}{dt}\right)_A > \left(\frac{dT}{dt}\right)_B$
So, A cools down at faster rate.
BETA
