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JEE Main 20198 AprilMorning Shift
Exam
JEE Main
Year
2019
Date
8 April
Shift
Morning Shift
Subject
Physics
Chapter
Thermal Properties of Matter

JEE Main 2019 · 8 April · Morning Shift · Physics · Thermal Properties of Matter

Two identical beakers A and B contain equal volumes of two...

Two identical beakers AA and BB contain equal volumes of two different liquids at 60°C60 ° C each and left to cool down. Liquid in AA has density of 8×102kgm−38 \times {10}^{2}kg{m}^{- 3} and specific heat of 2000Jkg−1K−12000Jk{g}^{- 1}{K}^{- 1} while the liquid in BB has density 103kgm−3{10}^{3}kg{m}^{- 3} and specific heat of 4000Jkg−1K−14000Jk{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)Two identical beakers A and B contain equal volumes of two different liquids at 60 ° C each and left to cool down.Two identical beakers A and B contain equal volumes of two different liquids at 60 ° C each and left to cool down.Two identical beakers A and B contain equal volumes of two different liquids at 60 ° C each and left to cool down.Two identical beakers A and B contain equal volumes of two different liquids at 60 ° C each and left to cool down.

Options

  1. AAccording to the assumptions, what must be the sole substrate in the respiration pathway?
  2. BAccording to the assumptions, what must be the sole substrate in the respiration pathway?
  3. CAccording to the assumptions, what must be the sole substrate in the respiration pathway?
  4. DAccording to the assumptions, what must be the sole substrate in the respiration pathway?

Correct answer

B. According to the assumptions, what must be the sole substrate in the respiration pathway?

Step-by-step solution

From Newton's law of cooling,
Rate of cooling: (−dTdt)=4\sigmaeAT03(T−T0)ms∝1ms\left(- \dfrac{dT}{dt}\right) = \dfrac{4\sigmaeA{T}_{0}^{3}\left(T - {T}_{0}\right)}{ms}∝ \dfrac{1}{ms}
∴ρA<ρB∴ {ρ}_{A} < {ρ}_{B}
mA<mB{m}_{A} < {m}_{B}
And sA<sB{s}_{A} < {s}_{B}
mAsA<mBsB{m}_{A}{s}_{A} < {m}_{B}{s}_{B}
At t=0,(−dTdt)A>(−dTdt)Bt = 0, {\left(- \dfrac{dT}{dt}\right)}_{A} > {\left(- \dfrac{dT}{dt}\right)}_{B}
Initially, the rate of cooling for AA will be higher than BB and temperature (T)( T ) vs time (t)( t ) graph will be a curve.
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