JEE Main 2020 — Physics Thermodynamics

2020integerhard

Three containers ${C}_{1},{C}_{2}$ and ${C}_{3}$ have water at different temperatures. The table below shows the final temperature $T$ when different amounts of water (given in liters) are taken from each container and mixed (assume no loss of heat during the process)

${C}_{1}$	${C}_{2}$	${C}_{3}$	$T$
$1l$	$2l$	$--$	${60}^{o}C$
$--$	$1l$	$2l$	${30}^{o}C$
$2l$	$--$	$1l$	${60}^{o}C$
$1l$	$1l$	$1l$	$\theta$

The value of $\theta$ (in ${}^{o}C$ to the nearest integer) is___________

Official previous-year question

Held on 8 Jan 2020 · Verified 6 Jul 2026.

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Solution

Let ${\theta }_{1}$ , ${\theta }_{2}$ and ${\theta }_{3}$ be the temperatures of water in three containers.

$1{\theta }_{1}+2{\theta }_{2}+0{\theta }_{3}=(1+2)60$

${\theta }_{1}+2{\theta }_{2}=180$ …. (1)

$0\times {\theta }_{1}+1\times {\theta }_{2}+2\times {\theta }_{3}=(1+2)30$

${\theta }_{2}+2{\theta }_{3}=90$ …… (2)

$2\times {\theta }_{1}+0\times {\theta }_{2}+1\times {\theta }_{3}=(2+1)60$

$2{\theta }_{1}+{\theta }_{3}=180$ …. (3)

and ${\theta }_{1}+{\theta }_{2}+{\theta }_{3}=(1+1+1)\theta$

from (1)+(2)+(3)

$3{\theta }_{1}+3{\theta }_{2}+3{\theta }_{3}=450\Longrightarrow {\theta }_{1}+{\theta }_{2}+{\theta }_{3}=150$

from (4) equation $150=3\theta \Longrightarrow \theta =50^{\circ}C$

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{C}_{1}

{C}_{2}

{C}_{3}

T

1l

2l

--

{60}^{o}C

--

1l

2l

{30}^{o}C

2l

--

1l

{60}^{o}C

1l

1l

1l

\theta

Solution

Let ${\theta }_{1}$ , ${\theta }_{2}$ and ${\theta }_{3}$ be the temperatures of water in three containers.

$1{\theta }_{1}+2{\theta }_{2}+0{\theta }_{3}=(1+2)60$

${\theta }_{1}+2{\theta }_{2}=180$ …. (1)

$0\times {\theta }_{1}+1\times {\theta }_{2}+2\times {\theta }_{3}=(1+2)30$

${\theta }_{2}+2{\theta }_{3}=90$ …… (2)

$2\times {\theta }_{1}+0\times {\theta }_{2}+1\times {\theta }_{3}=(2+1)60$

$2{\theta }_{1}+{\theta }_{3}=180$ …. (3)

and ${\theta }_{1}+{\theta }_{2}+{\theta }_{3}=(1+1+1)\theta$

from (1)+(2)+(3)

$3{\theta }_{1}+3{\theta }_{2}+3{\theta }_{3}=450\Longrightarrow {\theta }_{1}+{\theta }_{2}+{\theta }_{3}=150$

from (4) equation $150=3\theta \Longrightarrow \theta =50^{\circ}C$

JEE Main 2020 — Physics Thermodynamics

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JEE Main 2020 — Physics Thermodynamics

Did you get this right?

Solution

Your note

JEE Main Thermodynamics in other years

More JEE Main Thermodynamics questions

Other Physics topics for JEE Main

Mechanics

Electromagnetism

Modern Physics

Optics

Waves & Oscillations

Keep practicing Thermodynamics