Question 1: Volume of Mixed Liquids

Solution

1. Find the Equilibrium Temperature ($\theta$)

Since the liquids are chemically the same, they have the same density $\rho$ (at reference temp) and specific heat capacity $s$. Mass is $m = \rho V$. Assuming $\gamma \Delta T \ll 1$, we can approximate masses as proportional to volumes.

Using the principle of calorimetry (Heat Lost = Heat Gained):

$$ m_1 s (\theta – \theta_1) + m_2 s (\theta – \theta_2) = 0 $$ $$ \rho V_1 (\theta – \theta_1) + \rho V_2 (\theta – \theta_2) \approx 0 $$

Solving for $\theta$:

$$ \theta (V_1 + V_2) = V_1 \theta_1 + V_2 \theta_2 \implies \theta = \frac{V_1 \theta_1 + V_2 \theta_2}{V_1 + V_2} $$

2. Find the Final Volume ($V$)

The final volume is the sum of the masses divided by the density at the new temperature $\theta$. Alternatively, we calculate the new volume of each component part expanding/contracting from their initial state to $\theta$.

$$ V = V_1 [1 + \gamma(\theta – \theta_1)] + V_2 [1 + \gamma(\theta – \theta_2)] $$ $$ V = (V_1 + V_2) + \gamma [ V_1(\theta – \theta_1) + V_2(\theta – \theta_2) ] $$

From the calorimetry equation derived in step 1, the term in the brackets $V_1(\theta – \theta_1) + V_2(\theta – \theta_2)$ is zero.

$$ V = (V_1 + V_2) + \gamma (0) $$