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Calorimetry

Question

The temperature of equal masses of three different liquids A, B and C are $12^{\circ} C$ , $19^{\circ} C$ and $28^{\circ} C$ respectively, the temperature when A and B are mixed is $16^{\circ} C$ , when B and C are mixed is $23^{\circ} C$ , what is the temperature when A and C are mixed?

Moderate

A

$28^{\circ} c$
B

$31^{\circ} c$
C

${20.26}^{\circ} c$
D

${19.5}^{\circ} c$

Solution

Case 1: When A and B mixed
Resultant= $16^{o} c$
apply principle of method of mixtures,
heat lost by hot body B= heat gained by cold body A
heat lost or gain = MS(temperature difference),
where M is mass of the body , S is specific heat of the body
$h e a t g a i n b y A = h e a t l o s t b y B, h e r e c o m m o n t e m p e r a t u r e i s 16^{0} C {MS}_{1} {[16-12]=MS}_{2} [19-16]$
$S_{2} = \frac{4}{3} S_{1}$ ----(1)
Case 2: When B and C mixed
heat gain by B = heat lost by C. here common temperature is 23 C
${MS}_{2} {[23-19]=MS}_{3} =[28-23]$
${4S}_{2} {=5S}_{3}$
substitute equation (1)
$4 \times \frac{4}{3} S_{1} {=5S}_{3}$
$S_{3} = \frac{16}{15} S_{1}$ ----(2)
Case 3: When A and C are mixed
heat gain by A = heat lost by C
${MS}_{1} {[T-12]=MS}_{3} [28-T]$
substitute equation (2)
$S_{1} [T-12]= \frac{16}{15} S_{1} [28-T]$
${T=20.26}^{o} c$
this is the temperature when A and C are mixed

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