The temperatures 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, it is 23°C. What will be the temperature when A and C are mixed?

#### Solution

Given:

Temperature of A = 12°C

Temperature of B = 19°C

Temperature of C = 28°C

Temperature of mixture of A and B = 16°C

Temperature of mixture of B and C = 23°C

Let the mass of the mixtures be *M *and the specific heat capacities of the liquids A, B and C be *C*_{A}, *C*_{B}, and *C*_{C}, respectively.

According to the principle of calorimetry, when A and B are mixed, we get

Heat gained by Liquid A = Heat lost by liquid B

⇒ *MC*_{A} (16 − 12) = *MC*_{B} (19 − 16)

⇒ 4*MC*_{A} = 3 *MC*_{B}

`rArrMC_A=(3/4)MC_B............(1)`

When B and C are mixed:

Heat gained by liquid B = Heat lost by liquid C

⇒*MC*_{B} (23 − 19) = *MC*_{C} (28 − 23)

⇒ 4*MC*_{B} = 5 *MC*_{C}

`rArr M_(C C)=(4/5)M_(C B).............(2)`

When A and C are mixed:

Let the temperature of the mixture be *T.* Then,

Heat gained by liquid A = Heat lost by liquid C

⇒ *MC*_{A} (*T* − 12) = *MC*_{C} (28 − *T*)

Using the values of *MC*_{A }and *MC*_{C}, we get

`rArr(3/4)MC_B(T-12)=(4/5)MC_B(28-T) ............["From eq. (1) and (2)"]`

`rArr(3/4)(T-12)=(4/5)(28-T)`

⇒(3×5) (T−12) = (4×4) (28−T)

⇒15T−180 = 448−16T

⇒31T = 628

`rArrT=628/31=20.253^oC`

⇒T = 20.3°C