# Three liquids a,b,c having specific heat 1, 0.5 , 0.25 respectively are at temperature 20 ,40 ,60 respectively find temperature equilibrium if all have the same mass?

Jun 29, 2017

See the other solution which is shorter and has correct steps.

31.4"^@C, rounded to one decimal place

#### Explanation:

Let us mix liquids $a \mathmr{and} b$ first.
Suppose their final temperature is ${T}^{\circ} C$

Using Law of Conservation of energy
Heat gained by $a =$ Heat Lost by $b$

$m {s}_{a} \left(T - 20\right) = m {s}_{b} \left(40 - T\right)$

Dividing both sides by $m$, inserting given values of specific heats, we get
$1 \times \left(T - 20\right) = 0.5 \times \left(40 - T\right)$
Multiplying both sides with $2$ we get

$2 T - 40 = 40 - T$
$\implies 3 T = 80$
=>T=80/3 "^@C

Let us add liquid $c$ to this mixture. Let the equilibrium temperature of this mixture be ${T}_{3}$

Using Law of Conservation of energy again
Heat gained by mixture of $a \mathmr{and} b =$ Heat Lost by $c$

$\left[m {s}_{a} \left({T}_{3} - \frac{80}{3}\right) + m {s}_{b} \left({T}_{3} - \frac{80}{3}\right)\right] = m {s}_{c} \left(60 - {T}_{3}\right)$

Dividing by $m$ and inserting given values of specific heats

$\left[1 \times \left({T}_{3} - \frac{80}{3}\right) + 0.5 \times \left({T}_{3} - \frac{80}{3}\right)\right] = 0.25 \times \left(60 - {T}_{3}\right)$

Multiply both sides with $4$ and simplify

$6 \times \left({T}_{3} - \frac{80}{3}\right) = \left(60 - {T}_{3}\right)$
$\implies 6 {T}_{3} + {T}_{3} = 60 + 160$
=>T_3=220/7=31.4"^@C, rounded to one decimal place

Jun 29, 2017

Let the equilibrium temperature of the mixture of three liquid (a,b.c) each of same mass $m$g be ${t}^{\circ} C$. ( assuming cgs system for calculation)
Given their respective sp.heat as ${s}_{a} = 1 , {s}_{b} = 0.5 \mathmr{and} {s}_{c} = 0.25$
By calorimetric principle system has not taken heat from surrounding.
So net heat gain is zero.
Hence

$m \times {s}_{a} \times \left(t - 20\right) + m \times {s}_{b} \times \left(t - 40\right) + m \times {s}_{c} \times \left(t - 60\right) = 0$

$\implies {s}_{a} \times \left(t - 20\right) + {s}_{b} \times \left(t - 40\right) + {s}_{c} \times \left(t - 60\right) = 0$

$\implies 1 \times \left(t - 20\right) + 0.5 \times \left(t - 40\right) + 0.25 \times \left(t - 60\right) = 0$

$\implies t - 20 + 0.5 t - 20 + 0.25 t - 15 = 0$

$\implies 1.75 t = 55$

$\implies t = \frac{55}{1.75} = \frac{220}{7} \approx {31.4}^{\circ} C$