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VITEEE 2018 Paper

Section: Physics

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Question : 22 of 125

Marks: +1, -0

The tempertaure of equal masses of three different liquids A, B and C are 12°C, 19°C and 28°C respectively. The temperature when A and B are mixed is 16°C and when B and C are mixed is 23°C . The temperature when A and C are mixed is

18.2°C
22°C
20.2°C
25.2°C

Solution:

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 is the temperature when A and C are mixed then,

C_A

(θ - 12) =

C_C

(28 - θ)

⇒

\frac{C_A}{C_C}

=

\frac{28-\theta}{\theta-12}

... (ii)

On solving equations (i) and (ii) θ = 20.2°C

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