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) ⇒= ...(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)
⇒=...(i)
Multiply Eq. (i) by Eq. (ii), we get
= 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) ⇒= ⇒= ⇒4T3−40=270−9T3 ⇒T3=23.84°C