Question

Two liquids A and B are at $32^{ \, o}C$ and $24^{ \, o}C.$ When mixed in equal masses, the temperature of mixture is found to be $28^{ \, o}C.$ Their specific heats are in the ratio

• A. $3:2$
• B. $2:3$
• C. $1:1$
• D. $4:3$

Answer 1

Answer: (C)

Let specific heats of liquids A and B are respectively $S_{A}$ and $S_{B}$ .
It is given that
$T_{A}=32^{ \, o}C, \, T_{B}=24^{ \, o}C$
and $T_{m i x}=28^{ \, o}C$
When these liquids are mixed, then heat lost by $A=$ heat gained by B
$\Rightarrow \, mS_{A}\Delta T=mS_{B}\Delta T^{′}$
$\Rightarrow \, \, mS_{A}\left(T_{A} - T_{m i x}\right)=mS_{B}\left(T_{m i x} - T_{B}\right)$
$\Rightarrow \, \, S_{A}\left(32 - 28\right)=S_{B}\left(28 - 24\right)$
$\Rightarrow \, \, S_{A}\times 4=S_{B}\times 4$
$\Rightarrow \, \, S_{A}:S_{B}=1:1$