The temperature of equal masses of three different liquids A, B and C are 12ºC, 19ºC and 28ºC respectively. The temperature when A and B are mixed is 16ºC and when B and C are mixed is 23ºC. The temperature when A and C are mixed is
Text Solution
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Heat gain = heat lost
C A (16 –12) = C B (19 – 16) ⇒ ⇒ $\frac{C_A}{C_B}$ = $\frac{3}{4}$
and C B (23 – 19) = C C (28 – 23) ⇒ ⇒ $\frac{C^{B}}{C^{c}}$ = $\frac{5}{4}$
⇒ ⇒ $\frac{C_{A}}{C_{c}} = \frac{15}{16}$ ...(i)
If θ θ is the temperature when A and C are mixed then,
$c_{A}(\theta - 12)$ = $c_c(28-\theta)$ ⇒ ⇒ $\frac{C_{A}}{C_{C}}$ $\frac{28-\theta}{\theta-12}$ ...(ii)
On solving equation (i) and (ii) θ θ = 20.2ºC.
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