Three liquids with masses m₁, m₂, m₃ are thoroughly mixed. If their specific heats are c₁ , c₂ , c₃ and their temperatures T₁, T₂, T₃ respectively, then the temperature of the mixture is

The correct option is: m₁c₁T₁ + m₂c₂T₂ + m₃c₃T₃/(m₁c₁+m₂c₂+m₃c₃)

Step-by-step Explanation:

To find the final temperature of the mixture when three liquids are thoroughly mixed, we apply the Principle of Calorimetry, which states that in an isolated system, the total heat lost by the hotter bodies must be equal to the total heat gained by the colder bodies. In other words, the net heat exchange is zero.

Let $T$ be the final equilibrium temperature of the mixture.

The heat exchange ($Q$) for any substance is given by the formula:
$Q=m·c·\Delta T$
where:
- $m$ is the mass of the liquid,
- $c$ is the specific heat capacity, and
- $\Delta T$ is the change in temperature.

For the three liquids, the heat exchange values are:
- For Liquid 1: $Q_1=m_1{c}_1(T-T_1)$
- For Liquid 2: $Q_2=m_2{c}_2(T-T_2)$
- For Liquid 3: $Q_3=m_3{c}_3(T-T_3)$

According to the principle of calorimetry, the sum of all heat exchanges in an isolated system is zero:
$Q_1+Q_2+Q_3=0$

Substituting the values of $Q_1$, $Q_2$, and $Q_3$ into the equation:
$m_1{c}_1(T-T_1)+m_2{c}_2(T-T_2)+m_3{c}_3(T-T_3)=0$

Expanding the terms:
$m_1{c}_{1}T-m_1{c}_1{T}_1+m_2{c}_{2}T-m_2{c}_2{T}_2+m_3{c}_{3}T-m_3{c}_3{T}_3=0$

Group all the terms containing $T$ on one side and the rest on the other side:
$(m_1{c}_1+m_2{c}_2+m_3{c}_3)T=m_1{c}_1{T}_1+m_2{c}_2{T}_2+m_3{c}_3{T}_3$

Now, solve for the equilibrium temperature $T$:
$T=\frac{m_1{c}_1{T}_1+m_2{c}_2{T}_2+m_3{c}_3{T}_3}{m_1{c}_1+m_2{c}_2+m_3{c}_3}$

This expression represents the final temperature of the mixture, which matches the second option when written as (m₁c₁T₁ + m₂c₂T₂ + m₃c₃T₃) / (m₁c₁ + m₂c₂ + m₃c₃).