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Thermal Properties of Matter

JEE Main level, numerical answer, medium

72 of 126

Question 72 of 126

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___________

Show answer and solution

Answer: 50

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}=\left(1+2\right)60

{\theta }_{1}+2{\theta }_{2}=180

…. (1)

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

{\theta }_{2}+2{\theta }_{3}=90

…… (2)

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

2{\theta }_{1}+{\theta }_{3}=180

…. (3)
and

{\theta }_{1}+{\theta }_{2}+{\theta }_{3}=\left(1+1+1\right)\theta

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

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

from (4) equation

150=3\theta ⟹\theta =50^{\circ}C

JEE Main question paper PDF

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JEE Main 2025 Apr 2 Shift 1 question paper

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