Question

Equal masses of three liquids $A,B$ and $C$have temperatures ${10}^{\circ }C,{25}^{\circ }C$ and ${40}^{\circ }C$ respectively. If A and B are mixed, the mixture has a temperature of ${15}^{\circ }C$. If $B$ and $C$ are mixed, the mixture has a temperature of ${30}^{\circ }C$. If $A$ and $C$ are mixed the temperature of the mixture is:

• A. ${29}^{\circ }C$
• B. ${25}^{\circ }C$
• C. ${20}^{\circ }C$
• D. ${16}^{\circ }C$

Answer 1

The correct option is D ${16}^{\circ }C$

By law of calorimetry, heat lost by one substance must be equal to that gained by the other when mixing.
$m\times C\times \Delta T=m^{\prime }\times C^{\prime }\times \Delta T^{\prime }$
Hence, as per the given information,

$C_A(15-10)=C_B(25-15)$

$C_A=2C_B$

Also, $C_B(30-25)=C_C(40-30)$

$C_B=2C_C$

Let temperature be T when A and C are mixed. Then,

$C_A(T-10)=C_C(40-T)$

$4C_c(T-10)=C_C(40-T)$

$4T-40=40-T$

$T={16}^{0}C$