Question

Take $C_{ice}=0.53cal/g{-}^{\circ }C$, $C_{water}=1.0cal/g{-}^{\circ }C$, $(L_f{)}_{water}=80cal/g$ and $(L_v{)}_{water}=529cal/g$. The temperatures of equal masses of three different liquids $A$, $B$ and $C$ are ${12}^{\circ }C$, ${19}^{\circ }C$ and ${28}^{\circ }C$ respectively. The temperature when $A$ and $B$ are mixed is ${16}^{\circ }C$ and when $B$ and $C$ are mixed is ${23}^{\circ }C$. What would be the temperature when $A$ and $C$ are mixed ?

• A. ${20.25}^{\circ }C$
• B. ${30.25}^{\circ }C$
• C. ${40.25}^{\circ }C$
• D. ${10.25}^{\circ }C$

Answer 1

The correct option is A ${20.25}^{\circ }C$

We know that when two fluids $A$ and $B$ are mixed, heat loss by $A$ is equal to that gained by $B$.

$m_A{C}_A(T-T_A)=m_B{C}_B(T_B-T)$

Since masses are equal,

$C_A(16-12)=C_B(19-16)$

$C_A\left(4\right)=C_B\left(3\right)$

$C_B=\frac{4}{3}{C}_A$

Similarly, $C_B\left(4\right)=C_C\left(5\right)$

$C_A=\frac{15}{16}{C}_C$

Also, $C_A(T-12)=C_C(28-T)$

Solving it gives $T={20.25}^{o}C$