JEE Main 26-Aug-2021 Shift 2 Solved Paper

Given masses of three liquids are equal,
i.e. mx=my=mz=m
Temperature of mixture of liquids x and y,T1=16°C
Temperature of mixture of liquids y and z,T2=26°C
While mixing the liquids x and y, the heat energy lost by liquid y
will be equal to the heat energy absorbed by liquid x.
Consider the specific heat capacity of liquid x is sx, of liquid y is sy
and of liquid z is sz.
Now, for liquid x and y, using principle of calorimetry,
Heat gained by liquid x = Heat lost by liquid y
msx(T1−10)=msy(20−T1)
⇒sx(16−10)=sy(20−16)
⇒

sx

sy

=

2

3

...(i)
Similarly, for liquid y and z, using principle of calorimetry,
Heat gained by liquid y = Heat lost by liquid z
msy(T2−20)=msz(30−T2)
⇒sy(26−20)=sz(30−26) ...(ii)
⇒

sy

sz

=

2

3

...(i)
Multiply Eq. (i) by Eq. (ii), we get

sx

sz

=

4

9

Consider the mixing of liquids x and z, let the temperature of mixture be T3.
Using principle of calorimetry for liquids x and z.
Heat gained by liquid x = Heat lost by liquid z
msx(T3−10)=msz(30−T3)
⇒sx(T3−10)=sz(30−T3)
⇒

sx

sz

=

30−T3

T3−10

⇒

4

9

=

30−T3

T3−10

⇒4T3−40=270−9T3
⇒T3=23.84°C