Ratio of Latent Heat to Specific Heat Capacity in Calorimetry Process

To find the ratio of the latent heat of vaporization ($L$) of the liquid to its specific heat capacity ($s$), we analyze the heat exchange process in two distinct stages.

Let $C$ be the heat capacity of the calorimeter.

Stage 1: Evaporation of the first $5\text{ g}$ of liquid

The liquid is initially at $T_i = 30^\circ\text{C}$ and the calorimeter is at $T_c = 110^\circ\text{C}$. Since the liquid is poured very slowly, it absorbs heat from the calorimeter, rises to its boiling point ($T_b = 80^\circ\text{C}$), and evaporates. This process lowers the temperature of the calorimeter to the boiling point of the liquid, $80^\circ\text{C}$.

• Heat lost by the calorimeter: $Q_{\text{lost, 1}} = C \cdot (110^\circ\text{C} - 80^\circ\text{C}) = 30C$

• Heat gained by $5\text{ g}$ of the liquid to heat up to $80^\circ\text{C}$ and vaporize completely: $Q_{\text{gained, 1}} = m_1 \cdot s \cdot (80^\circ\text{C} - 30^\circ\text{C}) + m_1 \cdot L$ $Q_{\text{gained, 1}} = 5 \cdot s \cdot 50 + 5 \cdot L = 250s + 5L$

By conservation of energy: $30C = 250s + 5L \quad \text{--- (Equation 1)}$

---

Stage 2: Addition of $80\text{ g}$ of liquid

After the first stage, the calorimeter is at $80^\circ\text{C}$. An additional $80\text{ g}$ of liquid at $30^\circ\text{C}$ is added, bringing the final equilibrium temperature to $T_f = 50^\circ\text{C}$.

• Heat lost by the calorimeter: $Q_{\text{lost, 2}} = C \cdot (80^\circ\text{C} - 50^\circ\text{C}) = 30C$

• Heat gained by the $80\text{ g}$ of liquid: $Q_{\text{gained, 2}} = m_2 \cdot s \cdot (50^\circ\text{C} - 30^\circ\text{C})$ $Q_{\text{gained, 2}} = 80 \cdot s \cdot 20 = 1600s$

By conservation of energy: $30C = 1600s \quad \text{--- (Equation 2)}$

---

Solving for $\frac{L}{s}$

Equating Equation 1 and Equation 2: $1600s = 250s + 5L$

$1350s = 5L$

$\frac{L}{s} = \frac{1350}{5} = 270^\circ\text{C}$

Answer: 270