Question

Two liquids are at ${40}^{o}C$ and ${30}^{o}C.$ When they are mixed in equal masses , the temperature of the mixture is ${36}^{o}C.$ Ratio of their specific heats is

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

Answer 1

Answer: (A)

$3:2$

Let liquid - 1 has specific heat $s_1$ at temp.${40}^{\circ }C$

Liquid - 2 has specific heat $s_2$ at temp. ${30}^{\circ }C$

common temp. is ${36}^{\circ }C$ and both have equal mass (suppose $m$)

heat lost by liquid-1 is

$Q_1=m{s}_1\Delta t=m{s}_1(40-36)=4m{s}_1$

heat gained by liquid - 2 is $Q_2=m{s}_2\Delta t^{{}^{\prime }}$

$\Rightarrow Q_2=m{s}_2\times (36-30)=6m{s}_2$

both should be same according to the conservation of energy

$Q_1=Q_2$

or $4m{s}_1=6m{s}_2$

$\Rightarrow \frac{S_1}{S_2}=\frac{6}{4}=\frac{3}{2}$

$S_1{S}_2=3:2$