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Question: Two tanks A and B contain water at \({30^0}C{\kern 1pt} {\kern 1pt} {\kern 1pt} and{\kern 1pt} {\ker...

Two tanks A and B contain water at 300Cand800C{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 40Kg40Kg of water at 500C.{50^0}C.
A. 24Kg,16Kg24Kg,16Kg
B. 16Kg,24Kg16Kg,24Kg
C. 20Kg,20Kg20Kg,20Kg
D. 30Kg,10Kg30Kg,10Kg

Explanation

Solution

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 40Kg40Kg of water at 500C.{50^0}C.

Formula Used:
If mm is the mass of water and ΔT\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=mcΔTQ = 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 TA=300C{T_A} = {30^0}C and we need to prepare water at 500C{50^0}C so it is obvious when we take xx kilogram of water from tank a heat will be gained by water and that heat can be written as
Qgained(A)=mcΔT{Q_{gained(A)}} = mc\Delta T
Qgained(A)=xc(5030)\Rightarrow {Q_{gained(A)}} = xc(50 - 30)
Qgained(A)=20xc(i)\Rightarrow {Q_{gained(A)}} = 20xc \to (i)
Now let us assume we take yKgy\,Kg of water from tank B which is at temperature of 800C{80^0}C and we need to prepare water at 500C{50^0}C so it is obvious that heat will be lost so it can be written as
Qlost(B)=mcΔT{Q_{lost(B)}} = mc\Delta T
Qlost(B)=yc(8050)\Rightarrow {Q_{lost(B)}} = yc(80 - 50)
Qlost(B)=30yc(i)\Rightarrow {Q_{lost(B)}} = 30yc \to (i)

Now from principle of Heat gained == Heat lost we have,
Qgained(A)=Qlost(B){Q_{gained(A)}} = {Q_{lost(B)}}
20xc=30yc\Rightarrow 20xc = 30yc
x=32y(iii)\Rightarrow x = \dfrac{3}{2}y \to (iii)
Now, as total mass of water will be x+y=40(iv)x + y = 40 \to (iv)
So, solving equations for x and y put equation (iii) in to equation (IV) we get,
32y+y=40\dfrac{3}{2}y + y = 40
5y=80\Rightarrow 5y = 80
y=16Kg\Rightarrow y = 16Kg
From value of y we can find value of x using equation (iv)
x=4016x = 40 - 16
x=24Kg\therefore x = 24Kg
So, we need to take 24Kg24Kg of water from tank A and 16Kg16Kg 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.