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 CA, CB, and CC, 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
⇒ MCA (16 − 12) = MCB (19 − 16)
⇒ 4MCA = 3 MCB
`rArrMC_A=(3/4)MC_B............(1)`
When B and C are mixed:
Heat gained by liquid B = Heat lost by liquid C
⇒MCB (23 − 19) = MCC (28 − 23)
⇒ 4MCB = 5 MCC
`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
⇒ MCA (T − 12) = MCC (28 − T)
Using the values of MCA and MCC, 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
