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Two tanks $A$ and $B$ contain water at $30\,^oC$ and $80\,^oC$ respectively. Calculate the amount of water that must be taken from each tank to prepare $40\,kg$ water at $50\,^oC$
$24\,kg,\,16\,kg$
$16\,kg,\,24\,kg$
$20\,kg,\,20\,kg$
$30\,kg,\,10\,kg$
Solution
The amount of water that must be taken from $A$ and $B$ to prepare $40 k g$ water at $50^{\circ} \mathrm{C}$ is $24 \mathrm{kg}, 16 \mathrm{kg}(1)$
Let the mass of water taken from tank $A$ be $x$ kg.
Then, the mass of water taken from tank $\mathrm{B}=40-\mathrm{x} \mathrm{kg}$.
We know, $H=mst,$
where $m=m a s s$ of water, $s=$ specific heat of water and $t=temperature$ change.
Heat absorbed by $\tan \mathrm{k} \mathrm{A}=\mathrm{xs}(50-30) \mathrm{J}$
Heat released by tank $\mathrm{B}=(40-\mathrm{x}) \mathrm{s}(80-50) \mathrm{J}$
Now,
$H_{A}=H_{B}$
$\Rightarrow x s(50-30)=(40-x) s(80-50)$
$\Rightarrow 20 x=30(40-x)$
$\Rightarrow 2 x=120-3 x$
$\Rightarrow x=24$
$\cdot$ Mass of water taken from tank $\mathrm{A}=24 \mathrm{kg}$
$\cdot$ Mass of water taken from tank $\mathrm{B}=(40-24) \mathrm{kg}=16 \mathrm{kg}$
