Question

Two tanks A and B contain water at ${30^0}C{\kern 1pt} {\kern 1pt} {\kern 1pt} and{\kern 1pt} {\kern 1pt} {\kern 1pt} {80^0}C$ respectively, calculate the amount of water that must be taken from each tank to prepare $40Kg$ of water at ${50^0}C.$
A. $24Kg,16Kg$
B. $16Kg,24Kg$
C. $20Kg,20Kg$
D. $30Kg,10Kg$

Answer 1

Hint: In order to solve this question, we should know about basic concept of heat. Heat always flows from higher temperature body to lower temperature body so, a difference in temperature between two bodies makes them either heat lost or heat gain. Here, we will use the concept of heat gain and heat lost during preparing $40Kg$ of water at ${50^0}C.$

Formula Used:
If $m$ is the mass of water and $\Delta T$ be the temperature difference with another body and c be the heat capacity of water then , Heat energy either gained or lost by water is calculated as $Q = mc\Delta T$.
During the heat transfer between two bodies Heat gained $ = $ Heat lost.

Complete step by step answer:
It is given to us that in tank a water is at temperature of ${T_A} = {30^0}C$ and we need to prepare water at ${50^0}C$ so it is obvious when we take $x$ kilogram of water from tank a heat will be gained by water and that heat can be written as
${Q_{gained(A)}} = mc\Delta T$
$\Rightarrow {Q_{gained(A)}} = xc(50 - 30)$
$\Rightarrow {Q_{gained(A)}} = 20xc \to (i)$
Now let us assume we take $y\,Kg$ of water from tank B which is at temperature of ${80^0}C$ and we need to prepare water at ${50^0}C$ so it is obvious that heat will be lost so it can be written as
${Q_{lost(B)}} = mc\Delta T$
$\Rightarrow {Q_{lost(B)}} = yc(80 - 50)$
$\Rightarrow {Q_{lost(B)}} = 30yc \to (i)$

Now from principle of Heat gained $ = $ Heat lost we have,
${Q_{gained(A)}} = {Q_{lost(B)}}$
$\Rightarrow 20xc = 30yc$
$\Rightarrow x = \dfrac{3}{2}y \to (iii)$
Now, as total mass of water will be $x + y = 40 \to (iv)$
So, solving equations for x and y put equation (iii) in to equation (IV) we get,
$\dfrac{3}{2}y + y = 40$
$\Rightarrow 5y = 80$
$ \Rightarrow y = 16Kg$
From value of y we can find value of x using equation (iv)
$x = 40 - 16$
\[ \therefore x = 24Kg\]
So, we need to take $24Kg$ of water from tank A and $16Kg$ of water from tank B.

Hence, the correct option is A.

Note: It should be remembered that, loss of heat simply means that the temperature of body gets decreased while heat gain means the temperature of the body gets increased and heat capacity is the amount of heat needed to raise the temperature by one unit of a substance and it is fixed in magnitude for every material.