Question

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?

Answer 1

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)
⇒ 4MC_A = 3 MC_B
$\Rightarrow M{C}_{\mathrm{A}}=\left(\frac{3}{4}\right)M{C}_{\mathrm{B}}$ ...(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)
⇒ 4MC_B = 5 MC_C
$\Rightarrow {\mathrm{M}}_{\mathrm{CC}}=\left(\frac{4}{5}\right){\mathrm{M}}_{\mathrm{CB}}$ ...(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

$$\begin{gathered} \Rightarrow \left(\frac{3}{4}\right)M{C}_{\mathrm{B}}\left(T-12\right)=\left(\frac{4}{5}\right)M{C}_{\mathrm{B}}\left(28-T\right)\left[\mathrm{From}\mathrm{eqs}.\left(1\right)\mathrm{and}\left(2\right)\right] \\ \Rightarrow \left(\frac{3}{4}\right)\left(T-12\right)=\left(\frac{4}{5}\right)\left(28-T\right) \\ \Rightarrow \left(3\times 5\right)\left(T-12\right)=\left(4\times 4\right)\left(28-T\right) \\ \Rightarrow 15T-180=448-16T \\ \Rightarrow 31T=628 \\ \Rightarrow T=\frac{628}{31}=20.253°\mathrm{C} \\ \Rightarrow T=20.3°\mathrm{C} \end{gathered}$$