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?

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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?

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Answer 1 · 2017-06-29T05:12:54

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

$31.4^{\circ} C$ $31.4"^@C$ , rounded to one decimal place

Explanation:

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

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

$m s_{a} (T - 20) = m s_{b} (40 - T)$ $ms_a(T-20)=ms_b(40-T)$

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

$2 T - 40 = 40 - T$ $2T-40=40-T$
$\Rightarrow 3 T = 80$ $=>3T=80$
$\Rightarrow T = \frac{80}{3}^{\circ} C$ $=>T=80/3 "^@C$

Let us add liquid $c$ $c$ to this mixture. Let the equilibrium temperature of this mixture be $T_{3}$ $T_3$

Using Law of Conservation of energy again
Heat gained by mixture of $a and b =$ $aand b=$ Heat Lost by $c$ $c$

$[m s_{a} (T_{3} - \frac{80}{3}) + m s_{b} (T_{3} - \frac{80}{3})] = m s_{c} (60 - T_{3})$ $[ms_a(T_3-80/3)+ms_b(T_3-80/3)]=ms_c(60-T_3)$

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

$[1 \times (T_{3} - \frac{80}{3}) + 0.5 \times (T_{3} - \frac{80}{3})] = 0.25 \times (60 - T_{3})$ $[1xx(T_3-80/3)+0.5xx(T_3-80/3)]=0.25xx(60-T_3)$

Multiply both sides with $4$ $4$ and simplify

$6 \times (T_{3} - \frac{80}{3}) = (60 - T_{3})$ $6xx(T_3-80/3)=(60-T_3)$
$\Rightarrow 6 T_{3} + T_{3} = 60 + 160$ $=>6T_3+T_3=60+160$
$\Rightarrow T_{3} = \frac{220}{7} = 31.4^{\circ} C$ $=>T_3=220/7=31.4"^@C$ , rounded to one decimal place

Answer 2 · 2017-06-29T05:38:21

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

$m \times s_{a} \times (t - 20) + m \times s_{b} \times (t - 40) + m \times s_{c} \times (t - 60) = 0$ $mxxs_axx(t-20)+mxxs_bxx(t-40)+mxxs_cxx(t-60)=0$

$\Rightarrow s_{a} \times (t - 20) + s_{b} \times (t - 40) + s_{c} \times (t - 60) = 0$ $=>s_axx(t-20)+s_bxx(t-40)+s_cxx(t-60)=0$

$\Rightarrow 1 \times (t - 20) + 0.5 \times (t - 40) + 0.25 \times (t - 60) = 0$ $=>1xx(t-20)+0.5xx(t-40)+0.25xx(t-60)=0$

$\Rightarrow t - 20 + 0.5 t - 20 + 0.25 t - 15 = 0$ $=>t-20+0.5t-20+0.25t-15=0$

$\Rightarrow 1.75 t = 55$ $=>1.75t=55$

$\Rightarrow t = \frac{55}{1.75} = \frac{220}{7} \approx {31.4}^{\circ} C$ $=>t=55/1.75=220/7~~31.4^@C$

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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?

2 Answers

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