Question

The temperature 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

• A. 182 °C
• B. 22 °C
• C. 20.3 °C
• D. 24.2 °C

Answer 1

Answer: (C)

20.3 °C

Answer 2

Let $m_{1} = m_{2} = m_{3} = m$

Let$s_{1},s_{2},s_{3}$be the respective specific heats of the three liquids.

When A and B are mixed, temperature of mixture

$= 16{^\circ}C$

As heat gained by A = heat lost by B

$\therefore ms_{1}(16 - 12) = ms_{2}(19 - 16)$

$4s_{1} = 3s_{2}$ …. (i)

When B and C are mixed, temperature of mixture

$23{^\circ}C$

As heat gained by B = heat lost by C,

$ms_{2}(23 - 19) = ms_{3}(28 - 23)$

$\therefore$ $4s_{2} = 5s_{3}$ ….. (ii)

From (i) and (ii), $s_{1} = \frac{3}{4}s_{2} = \frac{15}{16}s_{3}$]

When A and C are mixed, suppose temperature of

mixture = t

Heat gained by A = Heat lost by C

$ms_{1}(t - 12) = s_{3}(28 - t)$

$\frac{15}{16}s_{3}(t - 12) = s_{3}(28 - t)$

15t – 180 = 448 – 16t

31 t = 448 + 180 = 628

$t = \frac{628}{31} = 20.3{^\circ}C$