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 × 102 kg/m3 and specific heat of 2000 J kg–1 K–1 while liquid in B has density of 103 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)

Heat and Thermodynamics PYQ — JEE Main 2019 · 8 Apr · Shift 1 · Q72 with Solution

View written solutionFree

Correct answer: B

Use Newton’s law of cooling

For a body cooling in surroundings at temperature $T_s$ ,

mc\,\frac{dT}{dt}=-hA(T-T_s)

where:

$m$ = mass of liquid,
$c$ = specific heat,
$A$ = effective surface area,
$h$ = heat transfer constant.

So,

\frac{dT}{dt}=-\frac{hA}{mc}(T-T_s)

Hence the cooling rate depends on the factor

\frac{1}{mc}

or equivalently on the thermal capacity $mc$ .

Since both beakers are identical and contain equal volumes, the liquid with larger $mc$ cools more slowly.

Compare heat capacities of the liquids

Given equal volume $V$ in both beakers,

m=\rho V

Therefore,

mc=\rho V c

Since $V$ is same for both, compare $\rho c$ .

For liquid A:

\rho_A=8\times 10^2\;\text{kg m}^{-3},\qquad c_A=2000\;\text{J kg}^{-1}\text{K}^{-1}

\rho_A c_A=(8\times 10^2)(2000)=1.6\times 10^6

For liquid B:

\rho_B=10^3\;\text{kg m}^{-3},\qquad c_B=4000\;\text{J kg}^{-1}\text{K}^{-1}

\rho_B c_B=(10^3)(4000)=4\times 10^6

Thus,

(\rho c)_B > (\rho c)_A

So,

(mc)_B > (mc)_A

Hence liquid B cools more slowly than liquid A.

Interpret the graph

Both start at the same initial temperature $60^\circ\text{C}$ .

Liquid A cools faster $\Rightarrow$ its temperature falls more steeply.
Liquid B cools slower $\Rightarrow$ its curve stays above A at later times.
Both curves approach room temperature asymptotically.

So the correct graph is the one where:

both start together at $60^\circ\text{C}$ ,
curve for A drops faster,
curve for B remains above A.

That corresponds to Option B.

Check with stored answer

Stored correct answer: B

My derived answer: B

So they agree.

More from Heat and Thermodynamics

View written solutionFree

Correct answer: B

Use Newton’s law of cooling

For a body cooling in surroundings at temperature $T_s$ ,

mc\,\frac{dT}{dt}=-hA(T-T_s)

where:

$m$ = mass of liquid,
$c$ = specific heat,
$A$ = effective surface area,
$h$ = heat transfer constant.

So,

\frac{dT}{dt}=-\frac{hA}{mc}(T-T_s)

Hence the cooling rate depends on the factor

\frac{1}{mc}

or equivalently on the thermal capacity $mc$ .

Since both beakers are identical and contain equal volumes, the liquid with larger $mc$ cools more slowly.

Compare heat capacities of the liquids

Given equal volume $V$ in both beakers,

m=\rho V

Therefore,

mc=\rho V c

Since $V$ is same for both, compare $\rho c$ .

For liquid A:

\rho_A=8\times 10^2\;\text{kg m}^{-3},\qquad c_A=2000\;\text{J kg}^{-1}\text{K}^{-1}

\rho_A c_A=(8\times 10^2)(2000)=1.6\times 10^6

For liquid B:

\rho_B=10^3\;\text{kg m}^{-3},\qquad c_B=4000\;\text{J kg}^{-1}\text{K}^{-1}

\rho_B c_B=(10^3)(4000)=4\times 10^6

Thus,

(\rho c)_B > (\rho c)_A

So,

(mc)_B > (mc)_A

Hence liquid B cools more slowly than liquid A.

Interpret the graph

Both start at the same initial temperature $60^\circ\text{C}$ .

Liquid A cools faster $\Rightarrow$ its temperature falls more steeply.
Liquid B cools slower $\Rightarrow$ its curve stays above A at later times.
Both curves approach room temperature asymptotically.

So the correct graph is the one where:

both start together at $60^\circ\text{C}$ ,
curve for A drops faster,
curve for B remains above A.

That corresponds to Option B.

Check with stored answer

Stored correct answer: B

My derived answer: B

So they agree.

Heat and Thermodynamics question

Heat and Thermodynamics question

For liquid A:

For liquid B:

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Heat and Thermodynamics question

Heat and Thermodynamics question

For liquid A:

For liquid B:

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