A liquid at 30°C is poured verly slowly into a Calorimeter that is at temperature of 110°C. The boiling temperature of the liquid is 80°C. It is found that the first 5 gm of the liquid completely evaporates. After pouring another 80 gm of the liquid the equilibrium temperature is found to be 50°C. The ratio of the Latent heat of the liquid to its specific heat will be ____C°.
[Neglect the heat exchange with surrounding.]

Explanation

To solve the problem, we will analyze the heat exchanges that occur when the liquid is poured into the calorimeter.

### Step-by-Step Solution:

1. **Understanding the Problem:**
- We have a liquid at 30°C being poured into a calorimeter at 110°C.
- The boiling point of the liquid is 80°C.
- The first 5 grams of the liquid completely evaporate.
- After pouring an additional 80 grams of the liquid, the equilibrium temperature is found to be 50°C.

2. **Heat Required for Evaporation:**
- The first 5 grams of the liquid must be heated from 30°C to 80°C (its boiling point) and then evaporated.
- The heat required to raise the temperature of the liquid from 30°C to 80°C:
\[ Q_1 = m \cdot c \cdot \Delta T = 5 \, \text{g} \cdot c \cdot (80 - 30) = 5c \cdot 50 = 250c \, \text{(calories)} \]
- The heat required for the evaporation of the liquid (latent heat):
\[ Q_2 = m \cdot L_v = 5 \cdot L_v \, \text{(calories)} \]

3. **Heat Lost by Calorimeter:**
- The calorimeter cools down from 110°C to 80°C while supplying heat to the 5 grams of liquid.
- The heat lost by the calorimeter:
\[ Q_3 = H \cdot \Delta T = H \cdot (110 - 80) = 30H \]
- By the conservation of energy, the heat gained by the liquid equals the heat lost by the calorimeter:
\[ Q_1 + Q_2 = Q_3 \]
- Substituting the values:
\[ 250c + 5L_v = 30H \quad \text{(Equation 1)} \]

4. **Heat Exchange with Additional 80 grams of Liquid:**
- The additional 80 grams of liquid is heated from 30°C to 50°C.
- The heat gained by the 80 grams of liquid:
\[ Q_4 = 80 \cdot c \cdot (50 - 30) = 80c \cdot 20 = 1600c \, \text{(calories)} \]
- The calorimeter cools from 80°C to 50°C:
\[ Q_5 = H \cdot (80 - 50) = 30H \]
- Again, by conservation of energy:
\[ Q_4 = Q_5 \]
- Substituting the values:
\[ 1600c = 30H \quad \text{(Equation 2)} \]

5. **Solving the Equations:**
- From Equation 2, we can express \(H\):
\[ H = \frac{1600c}{30} = \frac{160c}{3} \]
- Substitute \(H\) back into Equation 1:
\[ 250c + 5L_v = 30 \left(\frac{160c}{3}\right) \]
\[ 250c + 5L_v = 1600c \]
\[ 5L_v = 1600c - 250c \]
\[ 5L_v = 1350c \]
\[ L_v = \frac{1350c}{5} = 270c \]

6. **Finding the Ratio of Latent Heat to Specific Heat:**
- The ratio of the latent heat \(L_v\) to the specific heat \(c\):
\[ \frac{L_v}{c} = \frac{270c}{c} = 270 \]

### Final Answer:
The ratio of the latent heat of the liquid to its specific heat is **270°C**.