Question

two liquids are 40°c ,30°c when they are mixed in equal massess the temp of the mixture is 36°c rationof their specific heats is ​

Answer 1

Answer:

$\boxed{\mathfrak{Ratio \ of \ specific \ heats \ (s_1 : s_2) = 2:3}}$

Explanation:

According to the question both the liquid are of equal mass let it be 'm'

Let specific heat capacity of liquid at 40°C be $\rm s_1$ and specific heat capacity of liquid at 30°C be $\rm s_2$

Temperature of mixture (T) is given as 36°C

Heat lost by one liquid = Heat gained by other liquid

$\rm \implies \cancel{m}s_1(40 - T) = \cancel{m}s_2( T - 30) \\ \\ \rm \implies \dfrac{s_1}{s_2} = \dfrac{40 - T}{T - 30} \\ \\ \rm \implies \dfrac{s_1}{s_2} = \dfrac{40 - 36}{36 - 30} \\ \\ \rm \implies \dfrac{s_1}{s_2} = \dfrac{4}{6} \\ \\ \rm \implies \dfrac{s_1}{s_2} = \dfrac{2}{3} \\ \\ \rm \implies s_1 : s_2 = 2 : 3$

Answer 2

$\bf{\gray{\underbrace{\blue{GIVEN:-}}}}$

• Two liquids are 40°C and 30°C respectively .

• When they are mixed in equal masses the temperature of the mixture is 36°C .

$\bf{\gray{\underbrace{\blue{TO\:FIND:-}}}}$

• Ratio of their specific heat's .

$\bf{\gray{\underbrace{\blue{SOLUTION:-}}}}$

Let,

• Liquid-1 is 40°C .

• And Liquid-2 is 30°C .

Again,

• Liquid-1 has specific heat $\rm\red{s_1}$ at 40°C .

• And Liquid-2 has specific heat $\rm\red{s_2}$ at 30°C .

✍️ According to the question, common temperature is 36°C and

• Both have equal mass [Suppose m ] .

$\orange\bigstar\:\mathcal{\pink{\overbrace{\underbrace{\purple{Q\:=\:ms\:\triangle{t}\:}}}}}$

Where,

• m = mass

• s = specific heat

• ∆t = change in temperature .

FOR LIQUID-1 :-

$\green\bigstar\:\mathcal{\pink{\overbrace{\underbrace{\purple{Q_1\:=\:ms_1\:\triangle{t}\:}}}}}$

$\rm{\implies\:Q_1\:=\:ms_1\:(40\:-\:36)\:}$

$\rm\green{\implies\:Q_1\:=\:4ms_1\:}$

FOR LIQUID-2 :-

$\blue\bigstar\:\mathcal{\pink{\overbrace{\underbrace{\purple{Q_2\:=\:ms_2\:\triangle{t}\:}}}}}$

$\rm{\implies\:Q_2\:=\:ms_2\:(36\:-\:30)\:}$

$\rm\green{\implies\:Q_2\:=\:6ms_2\:}$

✨ Now, According to Conservation of energy,

• $\bf\pink{Q_1\:=\:Q_2}$

$\rm{\implies\:4ms_1\:=\:6ms_2\:}$

$\rm{\implies\:\dfrac{s_1}{s_2}\:=\:\dfrac{6m}{4m}\:}$

$\rm\orange{\implies\:\dfrac{s_1}{s_2}\:=\:\dfrac{3}{2}\:}$

$\rm\green{\implies\:s_1\::\:s_2\:=\:3\::\:2\:}$

$\rm\pink{\therefore}$ The ratio of their specific heat's is $\bf\purple{3\::\:2}$ .